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

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).

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 $(\mathbb Z,+)/n\mathbb Z$).

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 $(e_1,\ldots,e_r)$ 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).

Given: The objects and hypotheses in the example.

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).

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 $H$, $|G\times H|=|G|\,|H|$).

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

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 $1$. (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 $G=\langle a\rangle\oplus H.$ (A maximal-order cyclic subgroup splits off a finite abelian p-group).

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).

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).

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 $(\mathbb Z,+)/n\mathbb Z$).

Given: The objects and hypotheses in the example.

Let $n\ge1$ be an integer. Multiplication makes $\mathbb Z/n$ a commutative monoid with identity $[1]_n$ by thm-integers-modulo-n-basic-algebra. A class $u\in\mathbb Z/n$ is a unit when it is invertible in that monoid (def-invertible-element). The set of all units is $(\mathbb Z/n)^\times:=\{\,u\in\mathbb Z/n:\text{some }v\in\mathbb Z/n\text{ satisfies }uv=[1]_n\,\}.$ By lem-monoid-units-form-a-group, multiplication restricts to a group operation on $(\mathbb Z/n)^\times$, called the unit group modulo $n$. The quotient $\mathbb Z/n$ 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 $\varphi(n):=\big|(\mathbb Z/n)^\times\big|\in\mathbb N$ (def-finite-cardinality). For $n=1$, the quotient has one element, which is its multiplicative identity and hence a unit, so $\varphi(1)=1$ follows from the definition. (The unit group $(\mathbb{Z}/n)^\times$ and Euler's totient $\varphi(n)=\lvert(\mathbb{Z}/n)^\times\rvert$ for $n\ge1$).

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).

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).

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).

Given: The objects and hypotheses in the statement refuted.

On the set of pairs $(a,b)$ with $a, b \in \mathbb{Z}$ and $b \ne 0$, define $(a,b) \sim (c,d) \iff ad = cb \quad \text{in } \mathbb{Z}.$ This is an equivalence relation (lem-rat-equivalence). The rationals are the quotient $\mathbb{Q}$, and $[(a,b)]$ is written $a/b$. (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 $g^{0} = e, \qquad g^{\sigma(n)} = g^{n} \cdot g \quad (n \in \mathbb{N}).$ 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 $g^{0} = e$).

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 $X$, $\operatorname{Sym}(X) \;:=\; \{\, f : X \to X \;:\; f \text{ is a bijection} \,\},$ equipped with composition as its operation, $(f \circ g)(x) \;=\; f(g(x)) \qquad (x \in X),$ and with the identity map $\mathrm{id}_X$, given by $\mathrm{id}_X(x) = x$, as distinguished element. Composition of two bijections of $X$ is again a bijection of $X$ (def-injection-surjection-bijection), so $\operatorname{Sym}(X)$ is closed under $\circ$ and $\circ$ is a binary operation on it (def-binary-operation); and $\mathrm{id}_X$ is a bijection of $X$, so it is an element of $\operatorname{Sym}(X)$, and it is a two-sided identity for composition (def-identity-element) because $f \circ \mathrm{id}_X = f = \mathrm{id}_X \circ f$ holds pointwise for every $f$. That $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ 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 $(a\,b)$ with $a \ne b$: it exchanges $a$ and $b$ and fixes everything else, and it satisfies $(a\,b) \circ (a\,b) = \mathrm{id}_X$. A product of cycle symbols means their composite, so $(a\,b)(c\,d)$ is $(a\,b) \circ (c\,d)$. (The symmetric group $\operatorname{Sym}(X)$: the bijections of a set $X$ under composition).

For every set $X$, the triple $(\operatorname{Sym}(X), \circ, \mathrm{id}_X)$ 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 $\operatorname{Sym}(X)$ is not abelian: the transpositions $\tau = (a\,b)$ and $\rho = (b\,c)$ satisfy $\tau \circ \rho \ne \rho \circ \tau$. ($\operatorname{Sym}(X)$ 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 $n := \lvert A\rvert$ and write $\operatorname{Bij}(A) := \{\, f : A \to A \ :\ f \text{ is a bijection} \,\}.$ Then $\operatorname{Bij}(A)$ is finite and $\lvert\operatorname{Bij}(A)\rvert = n!$ (def-factorial-and-falling-factorial). More generally, for finite sets $X$ and $Y$ write $\operatorname{Bij}(X,Y)$ for the set of bijections $X \to Y$. If $\lvert X\rvert = \lvert Y\rvert = n$ then $\operatorname{Bij}(X,Y)$ is finite with $n!$ elements, and if $\lvert X\rvert \ne \lvert Y\rvert$ then $\operatorname{Bij}(X,Y) = \varnothing$. (A finite set $A$ with $\lvert A\rvert = n$ 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 $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ 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).

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 $(g,h)(g',h') := (gg',hh').$ 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 $(g,h)^{-1}=(g^{-1},h^{-1}).$ 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).