1.11: Commutative groups

The classification of finitely generated commutative groups is most naturally studied as part of the theory of modules over a principal ideal domain, but, for the sake of completeness, I include an elementary exposition here.

Let \(M\) be a commutative group, written additively. The subgroup \(\langle x_{1},\ldots,x_{k}\rangle\) of \(M\) generated by the elements \(x_{1}% ,\ldots,x_{k}\) consists of the sums \(\sum m_{i}x_{i}\), \(m_{i}\in\mathbb{Z}{}\). A subset \(\{x_{1},\ldots,x_{k}\}\) of \(M\) is a basis for \(M\) if it generates \(M\) and

\[m_{1}x_{1}+\cdots+m_{k}x_{k}=0,\quad m_{i}\in\mathbb{Z}{}\implies m_{i}% x_{i}=0\text{ for every }i; \nonumber \]

then

\[M=\langle x_{1}\rangle\oplus\cdots\oplus\langle x_{k}\rangle. \nonumber \]

[it19]Let \(x_{1},\ldots,x_{k}\) generate \(M\). For any \(c_{1},\ldots ,c_{k}\in\mathbb{N}{}\) with \(\gcd(c_{1},\ldots,c_{k})=1\), there exist generators \(y_{1},\ldots,y_{k}\) for \(M\) such that \(y_{1}=c_{1}x_{1}% +\cdots+c_{k}x_{k}\).

We argue by induction on \(s=c_{1}+\cdots+c_{k}\). The lemma certainly holds if \(s=1\), and so we assume \(s>1\). Then, at least two \(c_{i}\) are nonzero, say, \(c_{1}\geq c_{2}>0\). Now

• \(\{x_{1},x_{2}+x_{1},x_{3},\ldots,x_{k}\}\) generates \(M\),

• \(\gcd(c_{1}-c_{2},c_{2},c_{3},\ldots,c_{k})=1\), and

• \((c_{1}-c_{2})+c_{2}+\cdots+c_{k}<s\),

and so, by induction, there exist generators \(y_{1},\ldots,y_{k}\) for \(M\) such that

\[\begin{aligned} y_{1} & =(c_{1}-c_{2})x_{1}+c_{2}(x_{1}+x_{2})+c_{3}x_{3}+\cdots+c_{k}% x_{k}\\ & =c_{1}x_{1}+\cdots+c_{k}x_{k}\text{.}%\end{aligned} \nonumber \]

[it20] Every finitely generated commutative group \(M\) has a basis; hence it is a finite direct sum of cyclic groups.

⁹We argue by induction on the number of generators of \(M\). If \(M\) can be generated by one element, the statement is trivial, and so we may assume that it requires at least \(k>1\) generators. Among the generating sets \(\{x_{1},\ldots,x_{k}\}\) for \(M\) with \(k\) elements there is one for which the order of \(x_{1}\) is the smallest possible. We shall show that \(M\) is then the direct sum of \(\langle x_{1}\rangle\) and \(\langle x_{2},\ldots,x_{k}\rangle\). This will complete the proof, because the induction hypothesis provides us with a basis for the second group, which together with \(x_{1}\) forms a basis for \(M\).

If \(M\) is not the direct sum of \(\langle x_{1}\rangle\) and \(\langle x_{2},\ldots,x_{k}\rangle\), then there exists a relation

\[m_{1}x_{1}+m_{2}x_{2}+\cdots+m_{k}x_{k}=0\, \label{e16}% \]

with \(m_{1}x_{1}\neq0\). After possibly changing the sign of some of the \(x_{i}\), we may suppose that \(m_{1},\ldots,m_{k}\in\mathbb{N}{}\) and \(m_{1}<\mathrm{order}(x_{1})\). Let \(d=\gcd(m_{1},\ldots,m_{k})>0\), and let \(c_{i}=m_{i}/d\). According to the lemma, there exists a generating set \(y_{1},\ldots,y_{k}\) such that \(y_{1}=c_{1}x_{1}+\cdots+c_{k}x_{k}\). But

\[dy_{1}=m_{1}x_{1}+m_{2}x_{2}+\cdots+m_{k}x_{k}=0 \nonumber \]

and \(d\leq m_{1}<\text{order}(x_{1})\), and so this contradicts the choice of \(\{x_{1},\ldots,x_{k}\}\).

[it20a]A finite commutative group is cyclic if, for each \(n>0\), it contains at most \(n\) elements of order dividing \(n\).

After the Theorem [it20], we may suppose that \(G=C_{n_{1}% }\times\cdots\times C_{n_{r}}\) with \(n_{i}\in\mathbb{N}{}\). If \(n\) divides \(n_{i}\) and \(n_{j}\) with \(i\neq j\), then \(G\) has more than \(n\) elements of order dividing \(n\). Therefore, the hypothesis implies that the \(n_{i}\) are relatively prime. Let \(a_{i}\) generate the \(i\)th factor. Then \((a_{1}% ,\ldots,a_{r})\) has order \(n_{1}\cdots n_{r}\), and so generates \(G\).

[it20b]Let \(F\) be a field. The elements of order dividing \(n\) in \(F^{\times}\) are the roots of the polynomial \(X^{n}-1\). Because unique factorization holds in \(F[X]\), there are at most \(n\) of these, and so the corollary shows that every finite subgroup of \(F^{\times}\) is cyclic.

[it21] A nonzero finitely generated commutative group \(M\) can be expressed

\[M\approx C_{n_{1}}\times\cdots\times C_{n_{s}}\times C_{\infty}^{r} \label{e6}% \]

for certain integers \(n_{1},\ldots,n_{s}\geq2\) and \(r\geq0\). Moreover,

1. \(r\) is uniquely determined by \(M\);

2. the \(n_{i}\) can be chosen so that \(n_{1}\geq2\) and \(n_{1}|n_{2}% ,\ldots,n_{s-1}|n_{s}\), and then they are uniquely determined by \(M\);

3. the \(n_{i}\) can be chosen to be powers of prime numbers, and then they are uniquely determined by \(M\).

The number \(r\) is called the rank of \(M\). By \(r\) being uniquely determined by \(M\), we mean that in any two decompositions of \(M\) of the form ([e6]), the number of copies of \(C_{\infty}\) will be the same (and similarly for the \(n_{i}\) in (b) and (c)). The integers \(n_{1},\ldots,n_{s}\) in (b) are called the invariant factors of \(M\). Statement (c) says that \(M\) can be expressed

\[M\approx C_{p_{1}^{e_{1}}}\times\cdots\times C_{p_{t}^{e_{t}}}\times C_{\infty}^{r},\quad e_{i}\geq1, \label{e7}% \]

for certain prime powers \(p_{i}^{e_{i}}\) (repetitions of primes allowed), and that the integers \(p_{1}^{e_{1}},\ldots,p_{t}^{e_{t}}\) are uniquely determined by \(M\); they are called the elementary divisors of \(M\).

The first assertion is a restatement of Theorem [it20].

(a) For a prime \(p\) not dividing any of the \(n_{i}\),

\[M/pM\approx(C_{\infty}/pC_{\infty})^{r}\approx(\mathbb{Z}{}/p\mathbb{Z}{}% )^{r}, \nonumber \]

and so \(r\) is the dimension of \(M/pM\) as an \(\mathbb{F}{}_{p}\)-vector space.

(b,c) If \(\gcd(m,n)=1\), then \(C_{m}\times C_{n}\) contains an element of order \(mn\), and so

\[C_{m}\times C_{n}\approx C_{mn}. \label{e5}% \]

Use ([e5]) to decompose the \(C_{n_{i}}\) into products of cyclic groups of prime power order. Once this has been achieved, ([e5]) can be used to combine factors to achieve a decomposition as in (b); for example, \(C_{n_{s}% }=\prod C_{p_{i}^{e_{i}}}\), where the product is over the distinct primes among the \(p_{i}\) and \(e_{i}\) is the highest exponent for the prime \(p_{i}\).

In proving the uniqueness statements in (b) and (c), we can replace \(M\) with its torsion subgroup (and so assume \(r=0\)). A prime \(p\) will occur as one of the primes \(p_{i}\) in ([e7]) if and only \(M\) has an element of order \(p\), in which case \(p\) will occur exact \(a\) times, where \(p^{a}\) is the number of elements of order dividing \(p\). Similarly, \(p^{2}\) will divide some \(p_{i}^{e_{i}}\) in ([e7]) if and only if \(M\) has an element of order \(p^{2}\), in which case it will divide exactly \(b\) of the \(p_{i}^{e_{i}}\), where \(p^{a-b}p^{2b}\) is the number of elements in \(M\) of order dividing \(p^{2}\). Continuing in this fashion, we find that the elementary divisors of \(M\) can be read off from knowing the numbers of elements of \(M\) of each prime power order.

The uniqueness of the invariant factors can be derived from that of the elementary divisors, or it can be proved directly: \(n_{s}\) is the smallest integer \(>0\) such that \(n_{s}M=0\); \(n_{s-1}\) is the smallest integer \(>0\) such that \(n_{s-1}M\) is cyclic; \(n_{s-2}\) is the smallest integer such that \(n_{s-2}\) can be expressed as a product of two cyclic groups, and so on.

Each finite commutative group is isomorphic to exactly one of the groups

\[C_{n_{1}}\times\cdots\times C_{n_{r}},\quad n_{1}|n_{2},\ldots,n_{r-1}|n_{r}. \nonumber \]

The order of this group is \(n_{1}\cdots n_{r}\). For example, each commutative group of order \(90\) is isomorphic to exactly one of \(C_{90}\) or \(C_{3}\times C_{30}\) — to see this, note that the largest invariant factor must be a factor of \(90\) divisible by all the prime factors of \(90\).

The linear characters of a commutative group

Let \(\mu(\mathbb{C}{})=\{z\in\mathbb{C}{}\mid|z|=1\}\). This is an infinite group. For any integer \(n\), the set \(\mu_{n}(\mathbb{C}{})\) of elements of order dividing \(n\) is cyclic of order \(n\); in fact,

\[\mu_{n}(\mathbb{C}{})=\{e^{2\pi im/n}\mid0\leq m\leq n-1\}=\{1,\zeta ,\ldots,\zeta^{n-1},\} \nonumber \]

where \(\zeta=e^{2\pi i/n}\) is a primitive \(n\)th root of \(1\).

A linear character (or just character) of a group \(G\) is a homomorphism \(G\rightarrow\mu(\mathbb{C}{})\). The homomorphism \(a\mapsto1\) is called the trivial (or principal) character.

[it21a]The Legendre symbol modulo \(p\) of an integer \(a\) not divisible by \(p\) is

\[\left( \frac{a}{p}\right) \overset{\textup

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}{=}\left\{ \begin{array} [c]{rl}% 1 & \text{if }a\text{ is a square in }\mathbb{Z}/p\mathbb{Z}{}{}\\ -1 & \text{otherwise.}% \end{array} \right. \nonumber \]

Clearly, this depends only on \(a\) modulo \(p\), and if neither \(a\) nor \(b\) is divisible by \(p\), then \(\left( \frac{ab}{p}\right) =\left( \frac{a}% {p}\right) \left( \frac{b}{p}\right)\) (because \((\mathbb{Z}{}% /p\mathbb{Z}{})^{\times}\) is cyclic). Therefore \([a]\mapsto\left( \frac{a}% {p}\right) \colon(\mathbb{Z}{}/p\mathbb{Z}{})^{\times}\rightarrow\{\pm 1\}=\mu_{2}(\mathbb{C}{})\) is a character of \((\mathbb{Z}{}/p\mathbb{Z}% {})^{\times}\), sometimes called the quadratic character.

The set of characters of a group \(G\) becomes a group \(G^{\vee}\) under the addition,

\[(\chi+\chi^{\prime})(g)=\chi(g)\chi^{\prime}(g), \nonumber \]

called the dual group of \(G\). For example, the dual group \(\mathbb{Z}% {}^{\vee}\) of \(\mathbb{Z}{}\) is isomorphic to \(\mu(\mathbb{C}{})\) by the map \(\chi\mapsto\chi(1)\).

[it22]Let \(G\) be a finite commutative group.

1. The dual of \(G^{\vee}\) is isomorphic to \(G\).

2. The map \(G\rightarrow G^{\vee\vee}\) sending an element \(a\) of \(G\) to the character \(\chi\mapsto\chi(a)\) of \(G^{\vee}\) is an isomorphism.

In other words, \(G\approx G^{\vee}\) and \(G\simeq G^{\vee\vee}\).

The statements are obvious for cyclic groups, and \((G\times H)^{\vee}\simeq G^{\vee}\times H^{\vee}\).

[it23]The statement that the natural map \(G\rightarrow G^{\vee\vee}\) is an isomorphism is a special case of the Pontryagin theorem. For infinite groups, it is necessary to consider groups together with a topology. For example, as we observed above, \(\mathbb{Z}{}^{\vee}\simeq\mu(\mathbb{C}{})\). Each \(m\in\mathbb{Z}{}\) does define a character \(\zeta\mapsto\zeta^{m}% \colon\mu(\mathbb{C}{})\rightarrow\mu(\mathbb{C}{})\), but there are many homomorphisms \(\mu(\mathbb{C}{})\rightarrow\mu(\mathbb{C}{})\) not of this form, and so the dual of \(\mu(\mathbb{C}{})\) is larger than \(\mathbb{Z}{}\). However, these are the only continuous homomorphisms. In general, let \(G\) be a commutative group endowed with a locally compact topology¹⁰ for which the group operations are continuous; then the group \(G^{\vee}\) of continuous characters \(G\rightarrow\mu(\mathbb{C}{})\) has a natural topology for which it is locally compact, and the Pontryagin duality theorem says that the natural map \(G\rightarrow G^{\vee\vee}\) is an isomorphism.

[Orthogonality Relations][it24]Let \(G\) be a finite commutative group. For any characters \(\chi\) and \(\psi\) of \(G\),

\[\sum\nolimits_{a\in G}\chi(a)\psi(a^{-1})=\left\{ \begin{array} [c]{cc}% |G| & \text{if }\chi=\psi\\ 0 & \text{otherwise.}% \end{array} \right. \nonumber \]

In particular,

\[\sum\nolimits_{a\in G}\chi(a)=\left\{ \begin{array} [c]{cc}% |G| & \text{if }\chi\text{ is trivial}\\ 0 & \text{otherwise.}% \end{array} \right. \nonumber \]

If \(\chi=\psi\), then \(\chi(a)\psi(a^{-1})=1\), and so the sum is \(|G|\). Otherwise there exists a \(b\in G\) such that \(\chi(b)\neq\psi(b)\). As \(a\) runs over \(G\), so also does \(ab\), and so

\[\sum\nolimits_{a\in G}\chi(a)\psi(a^{-1})=\sum\nolimits_{a\in G}\chi (ab)\psi((ab)^{-1})=\chi(b)\psi(b)^{-1}\sum\nolimits_{a\in G}\chi (a)\psi(a^{-1}). \nonumber \]

Because \(\chi(b)\psi(b)^{-1}\neq1\), this implies that \(\sum\nolimits_{a\in G}\chi(a)\psi(a^{-1})=0\).

[it25]For any \(a\in G\),

\[\sum\nolimits_{\chi\in G^{\vee}}\chi(a)=\left\{ \begin{array} [c]{cc}% |G| & \text{if }a=e\\ 0 & \text{otherwise.}% \end{array} \right. \nonumber \]

Apply the theorem to \(G^{\vee}\), noting that \((G^{\vee})^{\vee }\simeq G\).