Finite Abelian Groups: Examples and Counterexamples

Given: The objects and hypotheses in the example.

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).

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 $\mathrm{\mid }G\mathrm{\mid }=n_1\cdots n_r$. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

Let $n_0,\dots ,n_{r-1}$ be a finite pairwise-coprime list of positive integers and let $N:={\prod }_{i<r}{n}_i$. The map $\mathrm{\Phi }:\mathbb{Z}\mathrm{/}N⟶{\prod }_{i<r}\mathbb{Z}\mathrm{/}{n}_i,[x{]}_N⟼([x{]}_{n_i}{)}_{i<r},$ 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).

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 $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}=(\mathbb{Z}\mathrm{/}n,+)$ 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 $(\mathbb{Z}\mathrm{/}n,+)$ is the quotient group $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}$).

Given: The objects and hypotheses in the example.

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).

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).

For $n>0$, a partition of $n$ is a finite nondecreasing list of positive integers $(e_1,\dots ,e_r)$ 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).

Given: The objects and hypotheses in the example.

Suppose $G\cong {\prod }_{j<r}{C}_{p^{e_j}}$ with $e_j\ge 1$, and in additive notation write $p^{i}G=\{p^{i}g:g\in G\}$. Define $d_i$ by $\mathrm{\mid }{p}^{i}G\mathrm{/}{p}^{i+1}G\mathrm{\mid }=p^{d_i}$. Then $d_i=\mathrm{\mid }\{j:e_j\ge i+1\}\mathrm{\mid }\mathrm{.}$ Consequently, for every $k\ge 1$, 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).

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

Given: The objects and hypotheses in the example.

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).

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 $\mathrm{\mid }G\mathrm{\mid }=n_1\cdots n_r$. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

Let $n\ge 1$, 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 $1$. (The number of finite abelian groups of order n is the product of the partition numbers of the prime exponents of n).

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).

Given: The objects and hypotheses in the example.

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 $G=⟨a⟩\oplus H\mathrm{.}$ (A maximal-order cyclic subgroup splits off a finite abelian p-group).

Let $G$ be a group and let $N_0,\dots ,N_{r-1}$ be normal subgroups, where $r\in \mathbb{N}$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap ⟨N_j:j<r,j\mathrm{\ne }i⟩=\{e\}\mathrm{.}$ 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).

Let $N_0,\dots ,N_{r-1}⊴G$. 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).

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 $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}=(\mathbb{Z}\mathrm{/}n,+)$ 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 $(\mathbb{Z}\mathrm{/}n,+)$ is the quotient group $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}$).

Given: The objects and hypotheses in the example.

Let $n\ge 1$ be an integer. Multiplication makes $\mathbb{Z}\mathrm{/}n$ a commutative monoid with identity $[1{]}_n$ by thm-integers-modulo-n-basic-algebra. A class $u\in \mathbb{Z}\mathrm{/}n$ is a unit when it is invertible in that monoid (def-invertible-element). The set of all units is $(\mathbb{Z}\mathrm{/}n{)}^{\times }:=\{u\in \mathbb{Z}\mathrm{/}n:\text{some }v\in \mathbb{Z}\mathrm{/}n\text{ satisfies }uv=[1{]}_n\}\mathrm{.}$ By lem-monoid-units-form-a-group, multiplication restricts to a group operation on $(\mathbb{Z}\mathrm{/}n{)}^{\times }$, called the unit group modulo $n$. The quotient $\mathbb{Z}\mathrm{/}n$ is finite with cardinality $n$ by thm-standard-representatives-modulo-n, and its unit set is a finite subset by thm-subset-of-a-finite-set. Euler's totient function is therefore defined for every positive integer $n$ by $\phi (n):=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid \in \mathbb{N}$ (def-finite-cardinality). For $n=1$, the quotient has one element, which is its multiplicative identity and hence a unit, so $\phi (1)=1$ follows from the definition. (The unit group $(\mathbb{Z}\mathrm{/}n{)}^{\times }$ and Euler's totient $\phi (n)=\mid (\mathbb{Z}\mathrm{/}n{)}^{\times }\mid$ for $n\ge 1$).

Let $G$ be a group and let $N_0,\dots ,N_{r-1}$ be normal subgroups, where $r\in \mathbb{N}$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap ⟨N_j:j<r,j\mathrm{\ne }i⟩=\{e\}\mathrm{.}$ 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).

Let $N_0,\dots ,N_{r-1}⊴G$. 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).

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 $\mathrm{\mid }G\mathrm{\mid }=n_1\cdots n_r$. The trivial group corresponds to the empty list and empty product. (Fundamental theorem of finite abelian groups: invariant-factor form).

Given: The objects and hypotheses in the statement refuted.

On the set of pairs $(a,b)$ with $a,b\in \mathbb{Z}$ and $b\mathrm{\ne }0$, define $(a,b)\sim (c,d)⟺ad=cb\text{in }\mathbb{Z}\mathrm{.}$ This is an equivalence relation (lem-rat-equivalence). The rationals are the quotient $\mathbb{Q}$, and $[(a,b)]$ is written $a\mathrm{/}b$. (The rationals as equivalence classes of pairs of integers).

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 $g^0=e,g^{\sigma (n)}=g^n\cdot g(n\in \mathbb{N})\mathrm{.}$ 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 $ng$ or $n\cdot g$ rather than $g^n$, with $0g=0$ and $\sigma (n)g=ng+g$; the definitions are identical, only the symbols differ. (Powers $g^n$: natural exponents in a monoid and integer exponents in a group, with $g^0=e$).

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).

Given: The objects and hypotheses in the statement.

Let $X$ be a set. A permutation of $X$ is a bijection $f:X\to X$ (def-injection-surjection-bijection). The symmetric group of $X$ is the set of all permutations of $X$, $\mathrm{Sym}(X):=\{f:X\to X:f\text{ is a bijection}\},$ equipped with composition as its operation, $(f\circ g)(x)=f(g(x))(x\in X),$ and with the identity map ${\mathrm{i}\mathrm{d}}_X$, given by ${\mathrm{i}\mathrm{d}}_X(x)=x$, as distinguished element. Composition of two bijections of $X$ is again a bijection of $X$ (def-injection-surjection-bijection), so $\mathrm{Sym}(X)$ is closed under $\circ$ and $\circ$ is a binary operation on it (def-binary-operation); and ${\mathrm{i}\mathrm{d}}_X$ is a bijection of $X$, so it is an element of $\mathrm{Sym}(X)$, and it is a two-sided identity for composition (def-identity-element) because $f\circ {\mathrm{i}\mathrm{d}}_X=f={\mathrm{i}\mathrm{d}}_X\circ f$ holds pointwise for every $f$. That $(\mathrm{Sym}(X),\circ ,{\mathrm{i}\mathrm{d}}_X)$ is a group is lem-symmetric-group-is-a-group. Cycle notation for a finite list of distinct points. For distinct elements $x_0,x_1,\dots ,x_{k-1}$ of $X$ with $k\ge 2$, the symbol $(x_0{x}_1\cdots x_{k-1})$ denotes the permutation sending $x_i$ to $x_{i+1}$ for $i<k-1$, sending $x_{k-1}$ to $x_0$, and fixing every element of $X$ outside $\{x_0,\dots ,x_{k-1}\}$. It is a bijection, because the map described sends the set $\{x_0,\dots ,x_{k-1}\}$ onto itself by a rule with an evident inverse (send each $x_{i+1}$ back to $x_i$ and $x_0$ back to $x_{k-1}$) and fixes the complement pointwise. A transposition is such a symbol with $k=2$, that is $(ab)$ with $a\mathrm{\ne }b$: it exchanges $a$ and $b$ and fixes everything else, and it satisfies $(ab)\circ (ab)={\mathrm{i}\mathrm{d}}_X$. A product of cycle symbols means their composite, so $(ab)(cd)$ is $(ab)\circ (cd)$. (The symmetric group $\mathrm{Sym}(X)$: the bijections of a set $X$ under composition).

For every set $X$, the triple $(\mathrm{Sym}(X),\circ ,{\mathrm{i}\mathrm{d}}_X)$ of def-symmetric-group is a group (def-group); the inverse of a permutation $f$ is its inverse function $f^{-1}$. If $X$ contains three distinct elements $a$, $b$, $c$, then $\mathrm{Sym}(X)$ is not abelian: the transpositions $\tau =(ab)$ and $\rho =(bc)$ satisfy $\tau \circ \rho \mathrm{\ne }\rho \circ \tau$. ($\mathrm{Sym}(X)$ is a group under composition, and it is non-abelian whenever $X$ has at least three distinct elements).

Let $A$ be a finite set with $n:=\mid A\mid$ and write $\mathrm{Bij}(A):=\{f:A\to A:f\text{ is a bijection}\}\mathrm{.}$ Then $\mathrm{Bij}(A)$ is finite and $\mid \mathrm{Bij}(A)\mid =n!$ (def-factorial-and-falling-factorial). More generally, for finite sets $X$ and $Y$ write $\mathrm{Bij}(X,Y)$ for the set of bijections $X\to Y$. If $\mid X\mid =\mid Y\mid =n$ then $\mathrm{Bij}(X,Y)$ is finite with $n!$ elements, and if $\mid X\mid \mathrm{\ne }\mid Y\mid$ then $\mathrm{Bij}(X,Y)=\mathrm{\varnothing }$. (A finite set $A$ with $\mid A\mid =n$ has exactly $n!$ bijections onto itself, and $n!$ bijections onto any set of the same cardinality).

Let $G$ be a group and $g\in G$, with integer powers as in def-group-power. Then $⟨g⟩=\{g^n:n\in \mathbb{Z}\},$ 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. ($⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian).

Let $G$ and $H$ be groups. Their external direct product has underlying set $G\times H:=\{(g,h):g\in G,h\in H\}$ and componentwise operation $(g,h)(g^{\mathrm{\prime }},h^{\mathrm{\prime }}):=(g{g}^{\mathrm{\prime }},h{h}^{\mathrm{\prime }})\mathrm{.}$ The fact that this operation makes $G\times H$ a group, with the indicated identity and inverses, is proved in thm-external-direct-product-is-a-group. Until that result is used, this definition introduces only the set and its componentwise binary operation. (The external direct product $G\times H$ with componentwise multiplication).

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 $(g,h{)}^{-1}=(g^{-1},h^{-1})\mathrm{.}$ 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).