Fundamental theorem of finite abelian groups: elementary-divisor form

Given: The objects and hypotheses in the statement.

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

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).

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).

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 $G=⟨g⟩$ 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 $n\ge 1$, then $G\cong (\mathbb{Z}\mathrm{/}n,+)$. (Every cyclic group is isomorphic to $(\mathbb{Z},+)$ or to $(\mathbb{Z}\mathrm{/}n,+)$ for its finite order $n\ge 1$).