The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group

Given: The objects and hypotheses in the statement.

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).

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).

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 $g^{0} = e$).

Let $G$ be a group and let $N\mathrel{\trianglelefteq}G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G/N$ has the left cosets $G/N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN.$ Independence of the chosen representatives is proved in thm-coset-multiplication-well-defined-iff-normal, and the group axioms are proved in thm-quotient-group-laws. (The quotient group $G/N$ and coset product $(gN)(hN)=ghN$).

Let $N\mathrel{\trianglelefteq}G$. If $[G:N]$ is finite, then the quotient group $G/N$ is finite and $|G/N|=[G:N].$ In particular, if $G$ is finite, then $|G/N|=\frac{|G|}{|N|}.$ (If $[G:N]$ is finite then $|G/N|=[G:N]$; for finite $G$ this equals $|G|/|N|$).

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

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