Abelian Groups

The structure theorem for finite Abelian groups (we shall not consider finitely generated infinite Abelian groups here) states that every finite Abelian group can be written as the direct sum of finite cyclic groups of prime power order $G \; \equiv \; \mathbb{Z}_{p_1^{a_1}} \oplus \mathbb{Z}_{p_2^{a_2}} \oplus \cdots \oplus \mathbb{Z}_{p_N^{a_N}}$ where $p_,p_2,\ldots,p_n$ are prime and $a_1,a_2,\ldots,a_N$ are positive integers. Moreover, the prime powers $p_1^{a_1},p_2^{a_2},\ldots,p_N^{a_N}$ are uniquely determined by $G$ to within ordering, and so determine $G$ to within isomorphism.

For a given finite Abelian group $G$ , let us tidy up this decomposition, sorting the cyclic groups into the different prime powers, and write $G \; \equiv \; \bigoplus_p \bigoplus_{n =1}^{A(p)} \mathbb{Z}_{p^{a(p)_n}}$ so that $G$ is constructed from $A(p)$ cyclic groups whose orders are powers of $p$ (for each prime $p$ ). Obviously, only a finite number of primes $p$ will actually contribute to this sum.

Let $X_G(p^n)$ be the number of elements of $G$ with order $p^n$ , and let $Y_G(p^n)$ be the number of elements of $G$ with order dividing $p^n$ , so that $Y_G(p^n) = \sum_{m=0}^n X_G(p^m)$ .

A cyclic group $C$ of order $p^k$ has $1$ element of order $1$ , $p-1$ elements of order $p$ , $p^2-p$ elements of order $p^2$ , $p^3 - p^2$ elements of order $p^3$ , ..., $p^k - p^{k-1}$ elements of order $p^k$ , so that $Y_C(p^m) \; =\; p^{\mathrm{min}(m,k)} \qquad \qquad m \ge 0 \;.$ Returning to our original group $G$ and its decomposition, we see that $Y_G(p^m) \; = \; \prod_{n=1}^{A(p)} p^{\mathrm{min}(m,a(p)_n)} \; = \; p^{\sum_{n=1}^{A(p)}\mathrm{min}(m,a(p)_n)} \qquad \qquad m \ge 0 \;.$

Thus, if we know the numbers of elements of each order in the group $G$ , we know all the numbers $Y_G(p^m)$ , and hence we know all the numbers $Z_{p,m} \; = \; \sum_{n=1}^{A(p)} \mathrm{min}(m,a(p)_n) \;, \qquad \qquad p\; \mathrm{ prime}, m \ge 0 \;.$ But $\begin{array}{rcl} \displaystyle Z_{p,1} & = & \displaystyle \sum_{n\,:\, a(p)_n \ge 1} a(p)_n \\ \displaystyle Z_{p,2} - Z_{p,1} & = & \displaystyle \sum_{n\,:\, a(p)_n \ge 2} a(p)_n \\ \displaystyle Z_{p,3} - Z_{p,2} & = & \displaystyle \sum_{n\,:\, a(p)_n \ge 3} a(p)_n \end{array}$ and so on. In other words, if we know all the numbers $Z_{p,m}$ , then we know (for each $p$ ) how many of the $a(p)_n$ are equal to $1$ , how many are equal to $2$ , how many are equal to $3$ , and so on. In other words, we know all the values of the $a(p)_n$ (to within ordering). But this means that we know the group $G$ to within isomorphism.

In summary, if we know how many elements of each order a finite Abelian group possesses (from the above, it is sufficient to know the number of elements of each prime power order), then the group is determined to within isomorphism. If two finite Abelian groups possess the same numbers of elements of each order, then the groups are isomorphic.