The Fundamental Theorem of Finite Abelian Groups

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Internal direct products of finitely many normal subgroups

Definition

Let $G$ be a group and let $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$ . They form an internal direct product when they generate $G$ and, for each $i<r$ , $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$ ; in additive notation one writes $G=H\oplus K$ . Normal subgroups and generated subgroups are those of Normal subgroup: invariance under conjugation and The subgroup $\langle S \rangle$ generated by a subset, the cyclic subgroup $\langle g \rangle$ , and cyclic groups, and the comparison product is The external direct product $G\times H$ with componentwise multiplication.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Internal direct products are external direct products, equivalently every element has a unique factorisation

Statement

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$ . The following are equivalent: the $N_i$ form an internal direct product of $G$ ; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$ ; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be a group and let $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$ . They form an internal direct product when they generate $G$ and, for each $i<r$ , $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$ ; in additive notation one writes $G=H\oplus K$ . Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

[L2]

For groups $G$ and $H$ , the componentwise operation of def-external-direct-product-of-groups makes $G\times H$ a group. Its identity is $(e_G,e_H)$ , and $(g,h)^{-1}=(g^{-1},h^{-1}).$ Moreover the coordinate maps $\pi_G(g,h)=g$ and $\pi_H(g,h)=h$ are group homomorphisms. ( $G\times H$ is a group with identity $(e_G,e_H)$ , coordinatewise inverses, and homomorphic coordinate projections).

[L3]

Let $(M,\cdot,e)$ and $(M',\cdot',e')$ be monoids (def-semigroup-and-monoid). A monoid homomorphism from $M$ to $M'$ is a function $f : M \to M'$ such that - (H1) $f(x \cdot y) = f(x) \cdot' f(y)$ for all $x, y \in M$ ; - (H2) $f(e) = e'$ . Let $G$ and $G'$ be groups (def-group). A group homomorphism from $G$ to $G'$ is a function $f : G \to G'$ satisfying (H1) alone: $f(xy) \;=\; f(x)\, f(y) \qquad \text{for all } x, y \in G .$ Condition (H2) is not imposed for groups because it follows: a group homomorphism automatically satisfies $f(e) = e'$ and $f(x^{-1}) = f(x)^{-1}$ (lem-group-homomorphism-basic-properties). For monoids it does not follow and must be assumed, which is why the two definitions differ. A homomorphism from a structure to itself is an endomorphism. The identity map of $M$ is a monoid homomorphism, and a composite of monoid homomorphisms is one, since $(g \circ f)(xy) = g(f(x)f(y)) = g(f(x))\,g(f(y))$ and $(g \circ f)(e) = g(e') = e''$ ; the same computation, without the second clause, shows a composite of group homomorphisms is a group homomorphism. (Monoid homomorphism and group homomorphism).

[L4]

A group homomorphism is injective if and only if its kernel is trivial. For a group homomorphism $f:G\to H$ , $f$ is injective exactly when $\ker f=\{e_G\}$ . (A group homomorphism is injective if and only if its kernel is trivial).

[L5]

If $H\le G$ and $N\mathrel{\trianglelefteq}G$ , then $HN$ is a subgroup and $H\cap N\mathrel{\trianglelefteq}H$ . Here $HN:=\{hn:h\in H,\ n\in N\}$ . (If $H\le G$ and $N\mathrel{\trianglelefteq}G$ , then $HN$ is a subgroup and $H\cap N\mathrel{\trianglelefteq}H$ ).

Proof

technique · direct

1.1

The internal intersection condition gives $N_i\cap N_j=\{e\}$ for $i\ne j$ . Unique factorisation gives the same conclusion, since an element of $N_i\cap N_j$ has expressions supported in either coordinate. In either case normality puts $[N_i,N_j]$ inside $N_i\cap N_j$ , so distinct factors commute and the multiplication map $\mu((n_i))=n_0\cdots n_{r-1}$ is a homomorphism.

givenL1L2L3

1.2

Conversely, suppose that $\mu$ is an isomorphism. Coordinate subgroups in the external product commute, so their images $N_i$ commute, and surjectivity says that the factors generate $G$ . If $x\in N_i\cap\langle N_j:j\ne i\rangle$ , the commuting factors express $x$ as an ordered product of elements from the other $N_j$ . The tuple supported at $i$ and this tuple supported away from $i$ have the same image, so injectivity gives $x=e$ . Hence the factors form an internal direct product.

givenL1L2L3L4L5

2.1

Under the internal-product condition, the image of $\mu$ is the subgroup generated by the factors, hence is all of $G$ . If $\mu((n_i))=e$ , then each $n_i$ is the inverse of a product of the other factors and so lies in $N_i\cap\langle N_j:j\ne i\rangle$ ; therefore every $n_i=e$ . Thus $\mu$ is an isomorphism.

step 1.1L1L4L5

2.2

Under unique factorisation, every element has exactly one preimage under the homomorphism $\mu$ . Thus $\mu$ is bijective and hence is an isomorphism.

step 1.1L3L4

3.1

For the empty family, each condition says that $G$ is trivial. For one factor, each says that $N_0=G$ , and the multiplication map is then the identity after identifying the one-fold product with $N_0$ .

givenL1L2∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

The p-primary component of an abelian group

Definition

Let $G$ be an abelian group and $p$ a prime. Its $p$ -primary component is $G(p)=\{g\in G:g^{p^k}=e\text{ for some }k\in\mathbb N\}.$ Thus the identity is included by $k=0$ . In additive notation, $G(p)=\{g:p^kg=0\text{ for some }k\in\mathbb N\}$ . Powers and element orders use Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ and The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity. No finiteness or maximality is part of the definition.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Cauchy's theorem for finite abelian groups

Statement

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $P$ be a property of naturals such that for every $n \in \mathbb{N}$ , if $P(m)$ holds for all $m < n$ then $P(n)$ . Then $P(n)$ holds for all $n \in \mathbb{N}$ . (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

[L2]

Let $G$ be a finite group and $H\le G$ . Then $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$ , $|H|$ divides $|G|$ . (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

[L3]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L4]

If $G$ is abelian and $N\mathrel{\trianglelefteq}G$ , then $G/N$ is abelian. (Every quotient group of an abelian group is abelian).

[L5]

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L6]

Let $G$ be a finite group such that the positive integer $|G|$ is prime. Then every $g\ne e$ has order $|G|$ , satisfies $\langle g\rangle=G$ , and hence generates $G$ . In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

[L7]

Let $G$ be a group, $g \in G$ , and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\operatorname{ord}(g) = n$ with $n \in \mathbb{N}$ , $n \ge 1$ . Then: 1. for every $k \in \mathbb{Z}$ , $g^{k} = e$ if and only if $k = qn$ for some $q \in \mathbb{Z}$ , that is, if and only if $n \mid k$ (thm-division-algorithm-in-z); 2. the powers $g^{0}, g^{1}, \dots, g^{n-1}$ are pairwise distinct: if $i, j \in \mathbb{N}$ with $i < n$ , $j < n$ and $g^{i} = g^{j}$ , then $i = j$ ; 3. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$ ; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$ . Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$ , $g^{j} = g^{k}$ implies $j = k$ ; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

[L8]

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ; - if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ . (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ ).

Proof

technique · induction

1.1

For strong induction on $|G|$ , the trivial group has no relevant prime divisor, and if $|G|=p$ then $G$ is cyclic of order $p$ .

basegivenL1L2L3L4L5L6L7L8

2.1

Fix the induction hypothesis for every finite abelian group of order smaller than $|G|$ . Choose $x\ne e$ . If $\langle x\rangle=G$ , cyclic-group classification supplies $x^{|G|/p}$ of order $p$ . Otherwise put $H=\langle x\rangle$ , a nontrivial proper subgroup.

ihstep 1.1

3.1

If $p\mid |H|$ , the induction hypothesis in $H$ gives an element of order $p$ . If $p\nmid |H|$ , then $p\mid |G/H|$ and the induction hypothesis in the smaller finite abelian quotient gives a coset $yH$ of order $p$ .

step 2.1

4.1

In the latter case $y^p\in H$ . Let $q$ be the order of $y^p$ ; then $q\mid |H|$ and $p\nmid q$ . Since the coset of $y$ has order $p$ , the order of $y$ is $pq$ , so $y^q$ has order $p$ .

step 3.1

5.1

Every branch supplies an element of order $p$ , completing the strong induction.

step 4.1discharge-induction∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A p-primary component has the full p-power order and is the unique subgroup of that order

Statement

Let $G$ be finite abelian and write $|G|=p^a m$ with $p\nmid m$ . Then $G(p)$ is a subgroup of order $p^a$ . It is the unique subgroup of $G$ having that order. In particular, if $p\nmid |G|$ , then $G(p)=\{e\}$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be an abelian group and $p$ a prime. Its $p$ -primary component is $G(p)=\{g\in G:g^{p^k}=e\text{ for some }k\in\mathbb N\}.$ Thus the identity is included by $k=0$ . In additive notation, $G(p)=\{g:p^kg=0\text{ for some }k\in\mathbb N\}$ . Powers and element orders use def-group-power and def-order-in-a-group. No finiteness or maximality is part of the definition. (The p-primary component of an abelian group).

[L2]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L3]

Let $G$ be a finite group and $H\le G$ . Then $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$ , $|H|$ divides $|G|$ . (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

[L4]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L5]

Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved. For $N\mathrel{\trianglelefteq}G$ , the maps $H\mapsto H/N$ and $K\mapsto\pi^{-1}(K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N\le H\le G$ and subgroups $K\le G/N$ ; they preserve normality. (Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L6]

Second isomorphism theorem for groups: $H/(H\cap N)\cong HN/N$ . If $H\le G$ and $N\mathrel{\trianglelefteq}G$ , then $H/(H\cap N)\cong HN/N.$ (Second isomorphism theorem for groups: $H/(H\cap N)\cong HN/N$ ).

[L7]

If $G$ is abelian and $N\mathrel{\trianglelefteq}G$ , then $G/N$ is abelian. (Every quotient group of an abelian group is abelian).

[L8]

Let $G$ be a group (def-group) with identity $e$ , let $g, h \in G$ , and let powers be as in def-group-power. For all $m, n \in \mathbb{Z}$ : 1. $g^{m+n} = g^{m} g^{n}$ ; 2. $g^{-m} = (g^{m})^{-1}$ ; 3. $(g^{m})^{n} = g^{mn}$ ; 4. $g^{m} g^{n} = g^{n} g^{m}$ : any two powers of one element commute; 5. if $gh = hg$ then $(gh)^{n} = g^{n} h^{n}$ . Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n = 2$ , and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $\mathbb{N}$ , and so does claim 5 for exponents in $\mathbb{N}$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

Proof

technique · direct

1.1

If $x^{p^r}=y^{p^s}=e$ and $t=\max\{r,s\}$ , commutativity and the power laws give $(xy)^{p^t}=e$ and $(x^{-1})^{p^t}=e$ , so $G(p)$ is a subgroup. If a prime $q$ divides $|G(p)|$ , Cauchy's theorem in $G(p)$ gives an element of order $q$ ; the definition of $G(p)$ forces $q=p$ . Hence $|G(p)|=p^b$ for some $b\le a$ .

givenL1L2L3L4L5L6L7L8

2.1

If $b<a$ , then $p$ divides $|G/G(p)|$ . Cauchy's theorem in this abelian quotient gives a nonidentity coset $xG(p)$ of order $p$ .

step 1.1

3.1

Then $x^p\in G(p)$ , so $(x^p)^{p^k}=e$ for some $k$ and therefore $x\in G(p)$ , contradicting the choice of a nonidentity coset. Hence $b=a$ .

step 2.1

4.1

If $H\le G$ has order $p^a$ , Lagrange applied inside $H$ makes every $h\in H$ have $p$ -power order, so $H\subseteq G(p)$ ; equal finite orders give $H=G(p)$ . The case $a=0$ gives the trivial subgroup.

step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A finite abelian group is the internal direct product of its primary components

Statement

If $G$ is finite abelian and $|G|=\prod_{i<r}p_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$ . Thus $G\cong\prod_{i<r}G(p_i).$ For the trivial group, this is the empty product.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be finite abelian and write $|G|=p^a m$ with $p\nmid m$ . Then $G(p)$ is a subgroup of order $p^a$ . It is the unique subgroup of $G$ having that order. In particular, if $p\nmid |G|$ , then $G(p)=\{e\}$ . (A p-primary component has the full p-power order and is the unique subgroup of that order).

[L2]

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$ . The following are equivalent: the $N_i$ form an internal direct product of $G$ ; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$ ; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L3]

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of lem-units-of-z. Call $p : r \to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i = p_j$ forces $i = j$ (def-injection-surjection-bijection). Let $n \in \mathbb{Z}$ with $n \ge 1$ and let $p : r \to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i < r$ . Then, with $v_q$ as in def-p-adic-valuation: 1. $\displaystyle n \;=\; \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; 2. $v_q(n) = 0$ for every prime $q$ that is not among $p_0,\dots,p_{r-1}$ ; 3. the exponents are determined by $n$ : if $e : r \to \mathbb{N}$ and $n = \prod_{i<r} p_i^{\,e_i}$ , then $e_j = v_{p_j}(n)$ for every $j < r$ . Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; the exponents are determined by $n$ , and $v_q(n) = 0$ for every prime $q$ outside the list).

[L4]

Let $a, b \in \mathbb{Z}$ , not both $0$ , and put $I \;:=\; \{\, ax + by \;:\; x, y \in \mathbb{Z} \,\} .$ Then $I$ contains a positive element, and its least positive element is $\gcd(a,b)$ (def-common-divisor-and-gcd). In particular there are integers $x_0, y_0$ with $a x_0 + b y_0 \;=\; \gcd(a,b),$ so the equation $ax + by = \gcd(a,b)$ is solvable in $\mathbb{Z}$ . (Bézout's identity: for integers $a, b$ not both zero, $\gcd(a,b)$ is the least positive element of $\{\, ax + by : x, y \in \mathbb{Z} \,\}$ ; in particular $ax + by = \gcd(a,b)$ has an integer solution).

[L5]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$ . (For finite groups $G$ and $H$ , $|G\times H|=|G|\,|H|$ ).

[L6]

Let $G$ be a finite group and $H\le G$ . Then $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$ , $|H|$ divides $|G|$ . (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

Each $G(p_i)$ has order $p_i^{a_i}$ , and the product of these orders is $|G|$ . Distinct primary components have trivial intersection because an element in both has order dividing powers of two distinct primes.

givenL1L2L3L4L5L6

2.1

Multiplication from the external product of the primary components to $G$ is injective: a tuple in its kernel would place each component in the intersection with the product of the others, whose order is both a power of $p_i$ and coprime to $p_i$ .

step 1.1

3.1

The external product has order $\prod_i p_i^{a_i}=|G|$ . Its injective multiplication map therefore has image of order $|G|$ and is surjective.

step 2.1

4.1

The internal-direct-product recognition theorem gives the displayed isomorphism. When $G$ is trivial the prime list is empty and both sides are the trivial group.

step 3.1∎

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A nontrivial finite abelian p-group with a unique subgroup of order p is cyclic

Statement

Let $G$ be a nontrivial finite abelian $p$ -group. If $G$ has exactly one subgroup of order $p$ , then $G$ is cyclic.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L2]

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L3]

If $G$ is abelian and $N\mathrel{\trianglelefteq}G$ , then $G/N$ is abelian. (Every quotient group of an abelian group is abelian).

[L4]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L5]

Let $G$ be a group and $g \in G$ , with integer powers as in def-group-power. Then $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$ . Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ( $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , and every cyclic group is abelian).

[L6]

Let $G$ be a group, $g \in G$ , and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\operatorname{ord}(g) = n$ with $n \in \mathbb{N}$ , $n \ge 1$ . Then: 1. for every $k \in \mathbb{Z}$ , $g^{k} = e$ if and only if $k = qn$ for some $q \in \mathbb{Z}$ , that is, if and only if $n \mid k$ (thm-division-algorithm-in-z); 2. the powers $g^{0}, g^{1}, \dots, g^{n-1}$ are pairwise distinct: if $i, j \in \mathbb{N}$ with $i < n$ , $j < n$ and $g^{i} = g^{j}$ , then $i = j$ ; 3. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$ ; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$ . Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$ , $g^{j} = g^{k}$ implies $j = k$ ; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · contradiction

1.1

Assume for contradiction that $G$ is not cyclic. Choose $a\in G$ of maximal order $p^m$ and put $A=\langle a\rangle$ , which is then proper.

assume-contragivenL1L2L3L4L5L6

2.1

Cauchy's theorem in $G/A$ gives $b+A$ of order $p$ . Thus $pb=sa$ in additive notation for some integer $s$ , while $b\notin A$ .

step 1.1

3.1

Maximality gives $p^m b=0$ , so $p^{m-1}sa=0$ . Since $a$ has order $p^m$ , the integer $s$ is divisible by $p$ , say $s=pt$ .

step 2.1

4.1

Then $c=b-ta$ is nonzero, lies outside $A$ , and satisfies $pc=0$ . Its order- $p$ subgroup differs from the unique order- $p$ subgroup inside $A$ , contradicting the hypothesis. Therefore $G$ is cyclic.

step 3.1discharge-contradiction∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A maximal-order cyclic subgroup splits off a finite abelian p-group

Statement

Let $G$ be a finite abelian $p$ -group and let $a\in G$ have maximal element order. Then there is a subgroup $H\le G$ such that $G=\langle a\rangle\oplus H.$

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be a nontrivial finite abelian $p$ -group. If $G$ has exactly one subgroup of order $p$ , then $G$ is cyclic. (A nontrivial finite abelian p-group with a unique subgroup of order p is cyclic).

[L2]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L3]

Let $P$ be a property of naturals such that for every $n \in \mathbb{N}$ , if $P(m)$ holds for all $m < n$ then $P(n)$ . Then $P(n)$ holds for all $n \in \mathbb{N}$ . (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

[L4]

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L5]

Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved. For $N\mathrel{\trianglelefteq}G$ , the maps $H\mapsto H/N$ and $K\mapsto\pi^{-1}(K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N\le H\le G$ and subgroups $K\le G/N$ ; they preserve normality. (Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

[L6]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L7]

Let $G$ be a group and $g \in G$ , with integer powers as in def-group-power. Then $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$ . Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ( $\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$ , and every cyclic group is abelian).

[L8]

Let $G$ be a group and let $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$ . They form an internal direct product when they generate $G$ and, for each $i<r$ , $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$ ; in additive notation one writes $G=H\oplus K$ . Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

[L9]

Let $G$ be a group, $g \in G$ , and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\operatorname{ord}(g) = n$ with $n \in \mathbb{N}$ , $n \ge 1$ . Then: 1. for every $k \in \mathbb{Z}$ , $g^{k} = e$ if and only if $k = qn$ for some $q \in \mathbb{Z}$ , that is, if and only if $n \mid k$ (thm-division-algorithm-in-z); 2. the powers $g^{0}, g^{1}, \dots, g^{n-1}$ are pairwise distinct: if $i, j \in \mathbb{N}$ with $i < n$ , $j < n$ and $g^{i} = g^{j}$ , then $i = j$ ; 3. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$ ; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$ . Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$ , $g^{j} = g^{k}$ implies $j = k$ ; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$ , the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$ ).

Proof

technique · induction

1.1

For induction on $|G|$ , the trivial and cyclic cases hold with the evident complement.

basegivenL1L2L3L4L5L6L7L8L9

2.1

Fix the induction hypothesis for smaller finite abelian $p$ -groups, assume $G$ is noncyclic, and put $A=\langle a\rangle$ .

ihstep 1.1

3.1

Write $|A|=p^m$ . If $a^r$ has order $p$ , then $a^{pr}=e$ , so the order characterisation gives $p^{m-1}\mid r$ ; hence $A$ has the unique order- $p$ subgroup $\langle a^{p^{m-1}}\rangle$ . The preceding lemma and Cauchy's theorem therefore give an order- $p$ subgroup $B$ of $G$ different from it, and $A\cap B=\{0\}$ .

step 2.1

4.1

In $G/B$ the image of $a$ has the same order as $a$ because $A\cap B=\{0\}$ . It is still of maximal order: if $y+B$ had order larger than $p^m$ , then $(y+B)^{p^m}\ne B$ , so $y^{p^m}\ne0$ and $y$ would have order larger than that of $a$ .

step 3.1

5.1

Induction in $G/B$ gives $G/B=(A+B)/B\oplus H'/B$ for some subgroup $H'\ge B$ . Pulling back yields $G=A+H'$ , while $A\cap H'\subseteq A\cap B=\{0\}$ .

step 4.1

6.1

Thus $H=H'$ is the required complement. The order- $p$ and one-factor boundaries are included in the cyclic case, completing the induction.

step 5.1discharge-induction∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Every finite abelian p-group is a direct product of cyclic p-groups

Statement

Every finite abelian $p$ -group is isomorphic to a finite direct product of cyclic groups of prime-power order. The trivial $p$ -group is the empty product.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $G$ be a finite abelian $p$ -group and let $a\in G$ have maximal element order. Then there is a subgroup $H\le G$ such that $G=\langle a\rangle\oplus H.$ (A maximal-order cyclic subgroup splits off a finite abelian p-group).

[L2]

Let $P$ be a property of naturals such that for every $n \in \mathbb{N}$ , if $P(m)$ holds for all $m < n$ then $P(n)$ . Then $P(n)$ holds for all $n \in \mathbb{N}$ . (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

[L3]

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$ . The following are equivalent: the $N_i$ form an internal direct product of $G$ ; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$ ; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ; - if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ . (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ ).

Proof

technique · induction

1.1

For induction on $|G|$ , the trivial group gives the empty product.

basegivenL1L2L3L4

2.1

Fix the induction hypothesis for smaller finite abelian $p$ -groups. If $G$ is nontrivial, choose $a$ of maximal order and split $G=\langle a\rangle\oplus H$ .

ihstep 1.1

3.1

The cyclic factor $\langle a\rangle$ has prime-power order. If $H$ is nontrivial then $|H|<|G|$ , so the induction hypothesis decomposes $H$ into cyclic $p$ -groups.

step 2.1

4.1

Concatenating that decomposition with $\langle a\rangle$ and applying internal-product recognition gives the asserted external direct product, completing the induction.

step 3.1discharge-induction∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Elementary-divisor data for a finite abelian group

Definition

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G\cong C_{q_0}\times\cdots\times C_{q_{r-1}},$ where every $q_i>1$ is a prime power. The unordered multiset of the $q_i$ , counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ and The external direct product $G\times H$ with componentwise multiplication. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group

Statement

Suppose $G\cong\prod_{j<r}C_{p^{e_j}}$ with $e_j\ge1$ , and in additive notation write $p^iG=\{p^ig:g\in G\}$ . Define $d_i$ by $|p^iG/p^{i+1}G|=p^{d_i}$ . Then $d_i=|\{j:e_j\ge i+1\}|.$ Consequently, for every $k\ge1$ , the number of summands of order $p^k$ is $d_{k-1}-d_k$ , so the elementary divisors are intrinsic. The restriction to $k\ge1$ is the whole content of the hypothesis $e_j\ge1$ : no summand has order $p^0=1$ , and $d_{-1}$ is not defined.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G\cong C_{q_0}\times\cdots\times C_{q_{r-1}},$ where every $q_i>1$ is a prime power. The unordered multiset of the $q_i$ , counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data. (Elementary-divisor data for a finite abelian group).

[L2]

Every finite abelian $p$ -group is isomorphic to a finite direct product of cyclic groups of prime-power order. The trivial $p$ -group is the empty product. (Every finite abelian p-group is a direct product of cyclic p-groups).

[L3]

Natural exponents, in a monoid. Let $(M,\cdot,e)$ be a monoid (def-semigroup-and-monoid) and $g \in M$ . By the recursion theorem (thm-recursion), applied with the set $M$ , the element $e$ and the function $x \mapsto x \cdot g$ from $M$ to $M$ , there is exactly one function $\mathbb{N} \to M$ , written $n \mapsto g^{n}$ , with $g^{0} = e, \qquad g^{\sigma(n)} = g^{n} \cdot g \quad (n \in \mathbb{N}).$ In particular $g^{0} = e$ for every $g$ , including $g = e$ , and $g^{1} = g^{\sigma(0)} = e \cdot g = g$ . Since $\mathbb{N}$ contains $0$ (def-natural-numbers), the exponent $0$ is a genuine value of the definition and not a separate convention. Integer exponents, in a group. Let $G$ be a group (def-group) and $g \in G$ . Write $\iota : \mathbb{N} \to \mathbb{Z}$ for the embedding $k = [(k,0)]$ of lem-nat-embeds-int, which is injective, preserves addition, multiplication and order, and has as image exactly the nonnegative integers. For $x \in \mathbb{Z}$ define - $g^{x} := g^{k}$ , the natural power, when $0 \le x$ and $x = k$ ; - $g^{x} := (g^{k})^{-1}$ when $x < 0$ and $-x = k$ . Why this is well defined. The order on $\mathbb{Z}$ is total and antisymmetric (thm-int-ordered-ring, def-int-order), so exactly one of $0 \le x$ and $x < 0$ holds and the two clauses never both apply. In the first clause $x$ is nonnegative, so $x = k$ for some $k \in \mathbb{N}$ , and $k$ is unique because $\iota$ is injective. In the second clause $x < 0$ gives $0 = x + (-x) < 0 + (-x) = -x$ by compatibility of the order with addition (thm-int-ordered-ring, def-int-operations), so $-x$ is a positive integer and again $-x = k$ for a unique $k$ . The inverse $(g^{k})^{-1}$ is a single determined element by lem-inverse-unique and def-invertible-element. Finally the two readings of $g^{k}$ , as a natural power and as an integer power, agree by construction, so no ambiguity is introduced. Abbreviation. In an exponent we write $k$ for the integer $k$ when a natural number $k$ is used where an integer is expected; this is unambiguous because $\iota$ is injective and preserves the arithmetic and the order, and because the two readings of $g^{k}$ agree as just noted. Additive notation. When the group is written additively the same object is written $n g$ or $n \cdot g$ rather than $g^{n}$ , with $0 g = 0$ and $\sigma(n) g = n g + g$ ; the definitions are identical, only the symbols differ. (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ ).

[L4]

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L5]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L6]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$ . (For finite groups $G$ and $H$ , $|G\times H|=|G|\,|H|$ ).

[L7]

For every $n\in\mathbb N$ , view $n$ as its canonical nonnegative integer and put $n\mathbb Z:=\{nk:k\in\mathbb Z\}$ . Then the left cosets of $n\mathbb Z$ in $(\mathbb Z,+)$ are exactly the congruence classes modulo $n$ , and coset addition is the published addition of congruence classes. Thus $(\mathbb Z,+)/n\mathbb Z=(\mathbb Z/n,+)$ as the same group on the same underlying set. This includes $n=0$ and $n=1$ . (For every $n\in\mathbb N$ , the congruence-class group $(\mathbb Z/n,+)$ is the quotient group $(\mathbb Z,+)/n\mathbb Z$ ).

Proof

technique · direct

1.1

On a cyclic factor $C_{p^e}$ , the quotient $p^iC_{p^e}/p^{i+1}C_{p^e}$ is trivial when $i\ge e$ and has order $p$ when $i<e$ .

givenL1L2L3L4L5L6L7

2.1

Taking direct products componentwise therefore gives $|p^iG/p^{i+1}G|=p^{d_i}$ with $d_i$ equal to the number of exponents at least $i+1$ .

step 1.1

3.1

The number of exponents equal to $k$ is the number at least $k$ minus the number at least $k+1$ , namely $d_{k-1}-d_k$ .

step 2.1

4.1

Each subgroup $p^iG$ and quotient $p^iG/p^{i+1}G$ is defined intrinsically, and the sequence terminates at zero, so these differences uniquely recover all summands.

step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Fundamental theorem of finite abelian groups: elementary-divisor form

Statement

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

If $G$ is finite abelian and $|G|=\prod_{i<r}p_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$ . Thus $G\cong\prod_{i<r}G(p_i).$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

[L2]

Every finite abelian $p$ -group is isomorphic to a finite direct product of cyclic groups of prime-power order. The trivial $p$ -group is the empty product. (Every finite abelian p-group is a direct product of cyclic p-groups).

[L3]

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G\cong C_{q_0}\times\cdots\times C_{q_{r-1}},$ where every $q_i>1$ is a prime power. The unordered multiset of the $q_i$ , counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data. (Elementary-divisor data for a finite abelian group).

[L4]

Suppose $G\cong\prod_{j<r}C_{p^{e_j}}$ with $e_j\ge1$ , and in additive notation write $p^iG=\{p^ig:g\in G\}$ . Define $d_i$ by $|p^iG/p^{i+1}G|=p^{d_i}$ . Then $d_i=|\{j:e_j\ge i+1\}|.$ Consequently, for every $k\ge1$ , the number of summands of order $p^k$ is $d_{k-1}-d_k$ , so the elementary divisors are intrinsic. (The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group).

[L5]

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ; - if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ . (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ ).

Proof

technique · direct

1.1

Primary decomposition separates $G$ into its intrinsic $p$ -primary components, and cyclic decomposition expresses each component as a product of cyclic $p$ -groups. This proves existence.

givenL1L2L3L4L5

2.1

For a fixed prime $p$ , the successive quotients $p^iG(p)/p^{i+1}G(p)$ recover the multiplicity of every cyclic order $p^k$ .

step 1.1

3.1

Doing this independently for each prime proves uniqueness of the multiset of elementary divisors. The assertion concerns factor isomorphism types, not uniqueness of internal complements.

step 2.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Invariant-factor data for a finite abelian group

Definition

An invariant-factor list for a finite abelian group $G$ is a finite list of integers $1<n_1\mid n_2\mid\cdots\mid n_r$ together with an isomorphism $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . The cyclic factors and product use Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ and The external direct product $G\times H$ with componentwise multiplication. Unit factors are omitted. The trivial group has the empty list.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Elementary divisors regroup uniquely into invariant factors

Statement

Every multiset of prime-power elementary divisors regroups in exactly one way into an invariant-factor list.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G\cong C_{q_0}\times\cdots\times C_{q_{r-1}},$ where every $q_i>1$ is a prime power. The unordered multiset of the $q_i$ , counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data. (Elementary-divisor data for a finite abelian group).

[L2]

An invariant-factor list for a finite abelian group $G$ is a finite list of integers $1<n_1\mid n_2\mid\cdots\mid n_r$ together with an isomorphism $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. Unit factors are omitted. The trivial group has the empty list. (Invariant-factor data for a finite abelian group).

[L3]

Let $n_0,\ldots,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:=\prod_{i<r}n_i$ . The map $\Phi:\mathbb Z/N\longrightarrow\prod_{i<r}\mathbb Z/n_i,\qquad[x]_N\longmapsto([x]_{n_i})_{i<r},$ is a bijection. It preserves addition, multiplication, $[0]$ , and $[1]$ componentwise. For the empty list, $N=1$ and both sides have one element. (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L4]

Let $\iota:\mathbb N\to\mathbb Z$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge1$ , then in the external direct product $\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).$ (If $g$ and $h$ have finite orders $m$ and $n$ , then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$ ).

[L5]

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ; - if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ . (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ ).

[L6]

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of lem-units-of-z. Call $p : r \to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i = p_j$ forces $i = j$ (def-injection-surjection-bijection). Let $n \in \mathbb{Z}$ with $n \ge 1$ and let $p : r \to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i < r$ . Then, with $v_q$ as in def-p-adic-valuation: 1. $\displaystyle n \;=\; \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; 2. $v_q(n) = 0$ for every prime $q$ that is not among $p_0,\dots,p_{r-1}$ ; 3. the exponents are determined by $n$ : if $e : r \to \mathbb{N}$ and $n = \prod_{i<r} p_i^{\,e_i}$ , then $e_j = v_{p_j}(n)$ for every $j < r$ . Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; the exponents are determined by $n$ , and $v_q(n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1

For each prime $p$ , sort its exponents increasingly. Left-pad the shorter prime lists with zeros until all have the same length, then multiply the prime powers columnwise to obtain $n_1,\ldots,n_r$ .

givenL1L2L3L4L5

2.1

The aligned exponents are nondecreasing, so $n_1\mid\cdots\mid n_r$ . The Chinese remainder theorem identifies each column product of coprime cyclic groups with $C_{n_i}$ .

step 1.1

3.1

Conversely, canonical prime factorisation of each $n_i$ recovers every padded exponent column and hence the original elementary divisors.

step 2.1L6

4.1

The empty multiset gives the empty list, so uniqueness includes the trivial group.

step 3.1∎

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Fundamental theorem of finite abelian groups: invariant-factor form

Statement

For every finite abelian group $G$ there is a unique list $1<n_1\mid\cdots\mid n_r$ such that $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . Moreover $|G|=n_1\cdots n_r$ . The trivial group corresponds to the empty list and empty product.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

[L2]

Every multiset of prime-power elementary divisors regroups in exactly one way into an invariant-factor list. (Elementary divisors regroup uniquely into invariant factors).

[L3]

An invariant-factor list for a finite abelian group $G$ is a finite list of integers $1<n_1\mid n_2\mid\cdots\mid n_r$ together with an isomorphism $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. Unit factors are omitted. The trivial group has the empty list. (Invariant-factor data for a finite abelian group).

[L4]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$ . (For finite groups $G$ and $H$ , $|G\times H|=|G|\,|H|$ ).

Proof

technique · direct

1.1

The elementary-divisor theorem supplies a unique multiset of prime powers, and the regrouping lemma converts it into a unique invariant-factor list.

givenL1L2L3L4

2.1

The order formula for finite direct products gives $|G|=\prod_i n_i$ ; for the empty list this product is $1$ , the order of the trivial group.

step 1.1∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Converse of Lagrange for finite abelian groups: every divisor occurs as a subgroup order

Statement

Let $G$ be finite abelian and let $d$ be a positive divisor of $|G|$ . Then $G$ has a subgroup of order $d$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Let $d, a \in \mathbb{Z}$ (def-integers). We say $d$ divides $a$ , and write $d \mid a$ , when $a \;=\; d q \qquad \text{for some } q \in \mathbb{Z},$ the product being that of def-int-operations. We write $d \nmid a$ when this fails. In this situation $d$ is called a divisor, or a factor, of $a$ , and $a$ is called a multiple of $d$ . This is the relation the library already has, not a second one. The published thm-division-algorithm-in-z introduces it in its own Statement, in these words: "We say $b$ divides $a$ , written $b \mid a$ , when $a = qb$ for some $q \in \mathbb{Z}$ ." Since multiplication on $\mathbb{Z}$ is commutative (thm-int-comm-ring), $a = qd$ and $a = dq$ are the same condition, so the definition above is that relation verbatim and the two usages agree everywhere. The theorem defined it for use on its own page and left the systematic theory to a later page; this is that page, and this item records the agreement rather than introducing a rival notion. The remainder test. For $b > 0$ the same Statement records that $b \mid a$ holds exactly when the remainder $r$ in $a = qb + r$ , $0 \le r < b$ , is $0$ . Boundary values. Each is one line from the ring axioms, and each is used below, so all three are recorded here rather than assumed: - $d \mid 0$ for every integer $d$ , including $d = 0$ , since $0 = d \cdot 0$ ; - $0 \mid a$ only for $a = 0$ , since $a = 0 \cdot q$ forces $a = 0$ ; - $1 \mid a$ and $a \mid a$ for every $a$ , since $a = 1 \cdot a$ and $a = a \cdot 1$ . (Divisibility in $\mathbb{Z}$ : $d \mid a$ when $a = dq$ for some integer $q$ ).

[L2]

The order of a finite group. Let $G$ be a group (def-group) whose underlying set is finite (def-countable), so that $G \approx n$ for some $n \in \mathbb{N}$ (def-equinumerous). That natural number is unique: if $G \approx n$ and $G \approx n'$ then $n \approx n'$ , since $\approx$ is symmetric and transitive, and then $n = n'$ by claim 3 of lem-pigeonhole. The order of $G$ is that unique natural number, written $|G|$ . A group is infinite when its underlying set is not finite, and $|G|$ is then not defined. The order of an element. Let $G$ be any group and $g \in G$ , with natural powers as in def-group-power. Put $S_g \;:=\; \{\, k \in \mathbb{N} \;:\; k \ge 1 \text{ and } g^{k} = e \,\} \;\subseteq\; \mathbb{N}.$ - If $S_g \ne \varnothing$ , the order of $g$ is its least element, $\operatorname{ord}(g) \;:=\; \min S_g \;\in\; \mathbb{N},$ which exists by the well-ordering principle (thm-well-ordering-principle): every nonempty subset of $\mathbb{N}$ has a least element, and that element is unique, being $\le$ every element of $S_g$ and a member of it. We then say $g$ has finite order. - If $S_g = \varnothing$ we say $g$ has infinite order and write $\operatorname{ord}(g) = \infty$ , where $\infty$ is a symbol reserved for this case and is not a natural number. No arithmetic is performed with it here. By construction $\operatorname{ord}(g) \ge 1$ whenever it is finite, and $\operatorname{ord}(g) = 1$ exactly when $g = e$ , since $g^{1} = g$ . Every element of a finite group has finite order. If $G$ is finite then $S_g \ne \varnothing$ for every $g \in G$ , by lem-order-of-element-exists, so $\operatorname{ord}(g)$ is a natural number. (The order $|G|$ of a finite group and the order $\operatorname{ord}(g)$ of an element, with $\operatorname{ord}(g) = \infty$ when no positive power of $g$ is the identity).

[L3]

Let $P$ be a property of naturals such that for every $n \in \mathbb{N}$ , if $P(m)$ holds for all $m < n$ then $P(n)$ . Then $P(n)$ holds for all $n \in \mathbb{N}$ . (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

[L4]

Let $n \in \mathbb{Z}$ with $n > 1$ , and put $S \;:=\; \{\, d \in \mathbb{Z} \;:\; d \mid n \ \text{ and } \ d > 1 \,\}$ (def-divides-in-z). Then $S$ is nonempty and has a least element $q$ , and $q$ is prime (def-prime). In particular every integer greater than $1$ has a prime divisor. (Every integer $n > 1$ has a prime divisor; indeed the least divisor of $n$ that exceeds $1$ is prime).

[L5]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L6]

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$ ).

[L7]

If $G$ is abelian and $N\mathrel{\trianglelefteq}G$ , then $G/N$ is abelian. (Every quotient group of an abelian group is abelian).

[L8]

Let $N\mathrel{\trianglelefteq}G$ . If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$ ; for finite $G$ this equals $|G|/|N|$ ).

[L9]

Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved. For $N\mathrel{\trianglelefteq}G$ , the maps $H\mapsto H/N$ and $K\mapsto\pi^{-1}(K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N\le H\le G$ and subgroups $K\le G/N$ ; they preserve normality. (Correspondence theorem: subgroups of $G/N$ correspond to subgroups of $G$ containing $N$ , with normality preserved).

Proof

technique · direct

1.1

Use strong induction on $|G|$ . If $d=1$ , take the trivial subgroup; this also settles the trivial group.

givenL1L2L3L4L5L6L7L8L9

2.1

For $d>1$ , choose a prime $p\mid d$ . Cauchy's theorem gives a subgroup $H\le G$ of order $p$ , and $G/H$ is finite abelian of order $|G|/p$ .

step 1.1

3.1

The integer $d/p$ divides $|G/H|$ , so induction gives a subgroup $K/H\le G/H$ of order $d/p$ .

step 2.1

4.1

By correspondence its full preimage $K\le G$ has $|K|=|H|\,|K/H|=d$ . The case $d=|G|$ returns $K=G$ .

step 3.1∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-11Open item page →

The exponent of a finite group

Definition

For a finite group $G$ , its exponent is $\exp(G)=\min\{n\in\mathbb N:n>0\text{ and }g^n=e\text{ for every }g\in G\}.$ The set is nonempty by $g^{|G|}=e$ for every element $g$ of a finite group $G$ , and The well-ordering principle gives its least member; powers use Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ . Thus the definition is well-defined. For the trivial group $\exp(G)=1$ .

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Invariant factors determine the order and exponent of a finite abelian group

Statement

If $G\cong C_{n_1}\times\cdots\times C_{n_r}$ with $1<n_1\mid\cdots\mid n_r$ , then $|G|=n_1\cdots n_r$ and $\exp(G)=n_r$ . For the empty list, $|G|=\exp(G)=1$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

For every finite abelian group $G$ there is a unique list $1<n_1\mid\cdots\mid n_r$ such that $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . Moreover $|G|=n_1\cdots n_r$ . The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

[L2]

For a finite group $G$ , its exponent is $\exp(G)=\min\{n\in\mathbb N:n>0\text{ and }g^n=e\text{ for every }g\in G\}.$ The set is nonempty by cor-g-to-the-group-order-is-identity, and thm-well-ordering-principle gives its least member; powers use def-group-power. Thus the definition is well-defined. For the trivial group $\exp(G)=1$ . (The exponent of a finite group).

[L3]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$ . (For finite groups $G$ and $H$ , $|G\times H|=|G|\,|H|$ ).

[L4]

Let $\iota:\mathbb N\to\mathbb Z$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge1$ , then in the external direct product $\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).$ (If $g$ and $h$ have finite orders $m$ and $n$ , then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$ ).

Proof

technique · direct

1.1

The finite-product order formula gives $|G|=\prod_i n_i$ , including the empty product.

givenL1L2L3L4

2.1

The order of an element of the product is the least common multiple of its component orders. Because $n_i\mid n_r$ , every element order divides $n_r$ , while an element generating the last factor has order $n_r$ .

step 1.1

3.1

The least common annihilating exponent is therefore $n_r$ when the list is nonempty, and is $1$ for the trivial group.

step 2.1∎

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

A nontrivial finite abelian group is cyclic if and only if it has one invariant factor

Statement

A nontrivial finite abelian group is cyclic if and only if its invariant-factor list has exactly one entry.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

For every finite abelian group $G$ there is a unique list $1<n_1\mid\cdots\mid n_r$ such that $G\cong C_{n_1}\times\cdots\times C_{n_r}$ . Moreover $|G|=n_1\cdots n_r$ . The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

[L2]

If $G\cong C_{n_1}\times\cdots\times C_{n_r}$ with $1<n_1\mid\cdots\mid n_r$ , then $|G|=n_1\cdots n_r$ and $\exp(G)=n_r$ . For the empty list, $|G|=\exp(G)=1$ . (Invariant factors determine the order and exponent of a finite abelian group).

[L3]

If $G=\langle g\rangle$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G\cong(\mathbb Z,+)$ ; - if $g$ has finite order $n$ , necessarily $n\ge1$ , then $G\cong(\mathbb Z/n,+)$ . (Every cyclic group is isomorphic to $(\mathbb Z,+)$ or to $(\mathbb Z/n,+)$ for its finite order $n\ge1$ ).

Proof

technique · direct

1.1

A one-entry invariant-factor decomposition is an isomorphism with one cyclic group, so $G$ is cyclic.

givenL1L2L3

2.1

Conversely a nontrivial finite cyclic group is isomorphic to $C_{|G|}$ , giving the one-entry list; uniqueness of invariant factors rules out any different list.

step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Indecomposable and decomposable nontrivial finite abelian groups

Definition

A nontrivial finite abelian group is indecomposable if it is not an internal direct product of two nontrivial subgroups in the sense of Internal direct products of finitely many normal subgroups. It is decomposable if such a product exists. The trivial group is assigned neither label.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Every nontrivial finite abelian group is an internal direct product of indecomposable subgroups

Statement

Every nontrivial finite abelian group is an internal direct product of finitely many indecomposable subgroups.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

A nontrivial finite abelian group is indecomposable if it is not an internal direct product of two nontrivial subgroups in the sense of def-internal-direct-product-of-subgroups. It is decomposable if such a product exists. The trivial group is assigned neither label. (Indecomposable and decomposable nontrivial finite abelian groups).

[L2]

Let $P$ be a property of naturals such that for every $n \in \mathbb{N}$ , if $P(m)$ holds for all $m < n$ then $P(n)$ . Then $P(n)$ holds for all $n \in \mathbb{N}$ . (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

[L3]

Let $N_0,\ldots,N_{r-1}\trianglelefteq G$ . The following are equivalent: the $N_i$ form an internal direct product of $G$ ; every $g\in G$ has a unique expression $g=n_0\cdots n_{r-1}$ with $n_i\in N_i$ ; and the multiplication map $\mu:\prod_{i<r}N_i\to G$ is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

Let $G$ be a finite group and $H\le G$ . Then $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$ , $|H|$ divides $|G|$ . (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · induction

1.1

For strong induction on $|G|$ , the order-one case is vacuous because the only group of that order is trivial.

basegivenL1L2L3L4

2.1

Fix a nontrivial $G$ and assume the result for every nontrivial finite abelian group of smaller order. If $G$ is indecomposable, the one-factor product is the required decomposition.

ihstep 1.1

3.1

If $G$ is decomposable, write $G=B\oplus C$ with $B$ and $C$ nontrivial. Lagrange gives $|B|,|C|<|G|$ , so the induction hypothesis decomposes each into indecomposable factors.

step 2.1

4.1

Unique factorisation in $B\oplus C$ and in the two inductive products combines to unique factorisation by all the smaller factors; internal-product recognition completes the induction.

step 3.1discharge-induction∎

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The indecomposable finite abelian groups are exactly the nontrivial cyclic groups of prime-power order

Statement

The indecomposable finite abelian groups are exactly the nontrivial cyclic groups of prime-power order.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

A nontrivial finite abelian group is indecomposable if it is not an internal direct product of two nontrivial subgroups in the sense of def-internal-direct-product-of-subgroups. It is decomposable if such a product exists. The trivial group is assigned neither label. (Indecomposable and decomposable nontrivial finite abelian groups).

[L2]

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

[L3]

Let $G$ be a finite abelian group and let $p$ be a prime dividing $|G|$ . Then $G$ contains an element, and hence a subgroup, of order $p$ . (Cauchy's theorem for finite abelian groups).

[L4]

Every subgroup $H$ of a cyclic group $G=\langle g\rangle$ is cyclic. If $H\ne\{e\}$ , then the least positive integer $d$ for which $g^d\in H$ satisfies $H=\langle g^d\rangle$ . (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).

[L5]

Let $G$ be a finite group such that the positive integer $|G|$ is prime. Then every $g\ne e$ has order $|G|$ , satisfies $\langle g\rangle=G$ , and hence generates $G$ . In particular, $G$ is cyclic. (A finite group of prime order is cyclic and every nonidentity element generates it).

Proof

technique · direct

1.1

The elementary-divisor theorem writes any nontrivial finite abelian group as a product of nontrivial cyclic prime-power factors. Indecomposability forces exactly one factor.

givenL1L2L3L4L5

2.1

Conversely, if $C_{p^a}=B\oplus C$ with both factors nontrivial, Cauchy's theorem gives an order- $p$ subgroup in each factor. Their images are distinct, but a cyclic group has a unique subgroup of each possible order. Hence no such decomposition exists.

step 1.1∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-11Open item page →

Partitions of a positive integer

Definition

For $n>0$ , a partition of $n$ is a finite nondecreasing list of positive integers $(e_1,\ldots,e_r)$ with $e_1+\cdots+e_r=n$ , using finite natural sums as in Finite sums and finite products of natural numbers, $\sum_{k<n} a_k$ and $\prod_{k<n} a_k$ in $\mathbb{N}$ and naturals as in The natural numbers $\mathbb{N}$ (von Neumann). Equality is equality of these lists. The nondecreasing convention removes permutations from the data.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Isomorphism classes of abelian groups of order p^n are counted by partitions of n

Statement

For a prime $p$ and $n>0$ , isomorphism classes of abelian groups of order $p^n$ are in bijection with partitions of $n$ . For $n=0$ , the unique group is the trivial group and corresponds separately to the empty partition.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

[L2]

For $n>0$ , a partition of $n$ is a finite nondecreasing list of positive integers $(e_1,\ldots,e_r)$ with $e_1+\cdots+e_r=n$ , using finite natural sums as in def-nat-finite-sum-and-product and naturals as in def-natural-numbers. Equality is equality of these lists. The nondecreasing convention removes permutations from the data. (Partitions of a positive integer).

[L3]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $|G\times H|=|G|\,|H|$ . (For finite groups $G$ and $H$ , $|G\times H|=|G|\,|H|$ ).

Proof

technique · direct

1.1

The elementary-divisor theorem writes such a group uniquely as $C_{p^{e_1}}\times\cdots\times C_{p^{e_r}}$ with the positive exponents arranged nondecreasingly. The product-order formula gives $e_1+\cdots+e_r=n$ .

givenL1L2L3

2.1

Thus the exponents form a partition of $n$ , and every partition constructs a group of order $p^n$ . Uniqueness of elementary divisors makes the two constructions inverse.

step 1.1∎

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

The number of finite abelian groups of order n is the product of the partition numbers of the prime exponents of n

Statement

Let $n\ge1$ , and write its canonical prime factorisation as $n=\prod_{i<r}p_i^{a_i}$ , with the $p_i$ distinct and $a_i>0$ . Then the number of isomorphism classes of abelian groups of order $n$ is $\prod_{i<r}P(a_i),$ where $P(a)$ is the number of partitions of $a$ . For $n=1$ one has $r=0$ , so the empty product is $1$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

If $G$ is finite abelian and $|G|=\prod_{i<r}p_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$ . Thus $G\cong\prod_{i<r}G(p_i).$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

[L2]

For a prime $p$ and $n>0$ , isomorphism classes of abelian groups of order $p^n$ are in bijection with partitions of $n$ . For $n=0$ , the unique group is the trivial group and corresponds separately to the empty partition. (Isomorphism classes of abelian groups of order p^n are counted by partitions of n).

[L3]

Powers are the natural powers of def-group-power and finite products those of def-monoid-finite-product, both taken in the commutative monoid $(\mathbb{Z},\cdot,1)$ of lem-units-of-z. Call $p : r \to \mathbb{Z}$ an injective list of primes when every $p_i$ is prime (def-prime) and $p_i = p_j$ forces $i = j$ (def-injection-surjection-bijection). Let $n \in \mathbb{Z}$ with $n \ge 1$ and let $p : r \to \mathbb{Z}$ be an injective list of primes such that every prime divisor of $n$ equals $p_i$ for some $i < r$ . Then, with $v_q$ as in def-p-adic-valuation: 1. $\displaystyle n \;=\; \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; 2. $v_q(n) = 0$ for every prime $q$ that is not among $p_0,\dots,p_{r-1}$ ; 3. the exponents are determined by $n$ : if $e : r \to \mathbb{N}$ and $n = \prod_{i<r} p_i^{\,e_i}$ , then $e_j = v_{p_j}(n)$ for every $j < r$ . Clause 3 needs only injectivity of the list, not the covering hypothesis. (For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i<r} p_i^{\,v_{p_i}(n)}$ ; the exponents are determined by $n$ , and $v_q(n) = 0$ for every prime $q$ outside the list).

[L4]

Let $(M,\cdot,e)$ be a monoid (def-semigroup-and-monoid) and let $g : \mathbb{N} \to M$ be a family of elements of $M$ , written $g_i := g(i)$ . There is exactly one function $P_g : \mathbb{N} \to M$ satisfying $P_g(0) = e, \qquad P_g(\sigma(n)) = P_g(n) \cdot g_n \quad (n \in \mathbb{N}),$ and we write $\prod_{i<n} g_i \;:=\; P_g(n), \qquad \text{also written } g_0 g_1 \cdots g_{n-1}.$ In particular the empty product is $\prod_{i<0} g_i = e$ , and $\prod_{i<1} g_i = e \cdot g_0 = g_0$ . Why the recursion is legitimate. The clause $P_g(\sigma(n)) = P_g(n) \cdot g_n$ consults $n$ as well as $P_g(n)$ , so thm-recursion does not apply to it directly. Apply that theorem instead with the set $A = \mathbb{N} \times M$ , the element $a = (0,e)$ , and the function $F : A \to A$ given by $F(n,x) = (\sigma(n),\, x \cdot g_n)$ : it yields a unique $H : \mathbb{N} \to \mathbb{N} \times M$ with $H(0) = (0,e)$ and $H(\sigma(n)) = F(H(n))$ . Writing $H(n) = (H_1(n), H_2(n))$ , induction (thm-induction-principle) gives $H_1(n) = n$ for every $n$ , since $H_1(0) = 0$ and $H_1(\sigma(n)) = \sigma(H_1(n))$ . Hence $H(\sigma(n)) = (\sigma(n),\, H_2(n) \cdot g_n)$ , so $P_g := H_2$ satisfies the two displayed equations. It is the only such function: if $Q$ satisfies them too, then $\{ n : P_g(n) = Q(n) \}$ contains $0$ and is closed under $\sigma$ , hence is all of $\mathbb{N}$ by induction. The value depends only on $g_0,\dots,g_{n-1}$ . If $g, g' : \mathbb{N} \to M$ satisfy $g_i = g'_i$ for every $i < n$ , then $P_g(n) = P_{g'}(n)$ . Indeed the set of $n$ for which this implication holds contains $0$ , both products then being $e$ ; and if it holds at $n$ , and $g, g'$ agree at every $i < \sigma(n)$ , then they agree at every $i < n$ and also at $n$ itself, because $i < \sigma(n)$ is equivalent to $i \le n$ (lem-nat-order-is-membership), so $P_g(\sigma(n)) = P_g(n) \cdot g_n = P_{g'}(n) \cdot g'_n = P_{g'}(\sigma(n))$ . Induction finishes it. This is what makes the notation $g_0 g_1 \cdots g_{n-1}$ unambiguous: it names a value determined by the first $n$ terms alone, and a finite list $u$ of length $n$ , that is a function $u : n \to M$ on the von Neumann natural $n = \{0,\dots,n-1\}$ (def-natural-numbers), determines the product $\prod_{i<n} u_i := P_{\tilde u}(n)$ computed from any extension $\tilde u : \mathbb{N} \to M$ of $u$ . (The product $g_0 g_1 \cdots g_{n-1}$ of a finite list in a monoid, by recursion, with the empty product ( $n = 0$ ) equal to the identity).

Proof

technique · direct

1.1

Primary decomposition makes an abelian group of order $n$ the product of one abelian $p_i$ -group of order $p_i^{a_i}$ for each $i<r$ .

givenL1L2L3L4

2.1

The choices for distinct primes are independent and the preceding corollary counts the $i$ th choice by $P(a_i)$ , so the product rule gives the formula. The empty prime factorisation of $1$ gives one choice.

step 1.1∎

DefinitionDefinition: AI-adaptedProof: Not applicableaudited 2026-08-11Open item page →

Squarefree positive integers

Definition

A positive integer $n$ is squarefree if no square of a prime divides $n$ . Equivalently, every exponent in its canonical prime factorisation is $0$ or $1$ . The integer $1$ is squarefree by the empty factorisation.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11Open item page →

Every abelian group of order n is cyclic if and only if n is squarefree

Statement

For a positive integer $n$ , every abelian group of order $n$ is cyclic if and only if $n$ is squarefree.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

A positive integer $n$ is squarefree if no square of a prime divides $n$ . Equivalently, every exponent in its canonical prime factorisation is $0$ or $1$ . The integer $1$ is squarefree by the empty factorisation. (Squarefree positive integers).

[L2]

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

[L3]

If $G$ is finite abelian and $|G|=\prod_{i<r}p_i^{a_i}$ is its prime factorisation, then the subgroups $G(p_i)$ form an internal direct product of $G$ . Thus $G\cong\prod_{i<r}G(p_i).$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

[L4]

Let $n_0,\ldots,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:=\prod_{i<r}n_i$ . The map $\Phi:\mathbb Z/N\longrightarrow\prod_{i<r}\mathbb Z/n_i,\qquad[x]_N\longmapsto([x]_{n_i})_{i<r},$ is a bijection. It preserves addition, multiplication, $[0]$ , and $[1]$ componentwise. For the empty list, $N=1$ and both sides have one element. (Chinese remainder theorem for a finite pairwise-coprime list: simultaneous residues determine one class modulo the product, and the resulting bijection preserves addition and multiplication).

[L5]

Let $\iota:\mathbb N\to\mathbb Z$ be the canonical embedding. If $g\in G$ and $h\in H$ have finite orders $m,n\ge1$ , then in the external direct product $\operatorname{ord}(g,h)=\operatorname{lcm}(m,n).$ (If $g$ and $h$ have finite orders $m$ and $n$ , then $\iota(\operatorname{ord}(g,h))=\operatorname{lcm}(\iota(m),\iota(n))$ in $G\times H$ ).

Proof

technique · direct

1.1

If $n$ is squarefree, each primary component of an abelian group of order $n$ has prime order and is cyclic. The Chinese remainder theorem combines the cyclic factors of pairwise coprime orders into a cyclic group of order $n$ .

givenL1L2L3L4L5

2.1

If $p^2\mid n$ , write $n=p^a m$ with $a\ge2$ and $(p,m)=1$ . The abelian group $C_p\times C_p\times C_{p^{a-2}}\times C_m$ , omitting trivial factors, has order $n$ but exponent strictly below $n$ , so it is not cyclic.

step 1.1

3.1

For $n=1$ the sole group is trivial and cyclic, agreeing with squarefreeness of $1$ .

step 2.1∎

The Fundamental Theorem of Finite Abelian Groups

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Internal direct products of finitely many normal subgroups

Definition

Internal direct products are external direct products, equivalently every element has a unique factorisation

Statement

Facts & Assumptions

Proof

The p-primary component of an abelian group

Definition

Cauchy's theorem for finite abelian groups

Statement

Facts & Assumptions

Proof

A p-primary component has the full p-power order and is the unique subgroup of that order

Statement

Facts & Assumptions

Proof

A finite abelian group is the internal direct product of its primary components

Statement

Facts & Assumptions

Proof

A nontrivial finite abelian p-group with a unique subgroup of order p is cyclic

Statement

Facts & Assumptions

Proof

A maximal-order cyclic subgroup splits off a finite abelian p-group

Statement

Facts & Assumptions

Proof

Every finite abelian p-group is a direct product of cyclic p-groups

Statement

Facts & Assumptions

Proof

Elementary-divisor data for a finite abelian group

Definition

The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group

Statement

Facts & Assumptions

Proof

Fundamental theorem of finite abelian groups: elementary-divisor form

Statement

Facts & Assumptions

Proof

Invariant-factor data for a finite abelian group

Definition

Elementary divisors regroup uniquely into invariant factors

Statement

Facts & Assumptions

Proof

Fundamental theorem of finite abelian groups: invariant-factor form

Statement

Facts & Assumptions

Proof

Converse of Lagrange for finite abelian groups: every divisor occurs as a subgroup order

Statement

Facts & Assumptions

Proof

The exponent of a finite group

Definition

Invariant factors determine the order and exponent of a finite abelian group

Statement

Facts & Assumptions

Proof

A nontrivial finite abelian group is cyclic if and only if it has one invariant factor

Statement

Facts & Assumptions

Proof

Indecomposable and decomposable nontrivial finite abelian groups

Definition

Every nontrivial finite abelian group is an internal direct product of indecomposable subgroups

Statement

Facts & Assumptions

Proof

The indecomposable finite abelian groups are exactly the nontrivial cyclic groups of prime-power order

Statement

Facts & Assumptions