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 $g^0=e,g^{\sigma (n)}=g^n\cdot g(n\in \mathbb{N})\mathrm{.}$ 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 $ng$ or $n\cdot g$ rather than $g^n$, with $0g=0$ and $\sigma (n)g=ng+g$; 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 $N⊴G$ be a normal subgroup (def-normal-subgroup). The quotient group, or factor group, $G\mathrm{/}N$ has the left cosets $G\mathrm{/}N:=\{gN:g\in G\}$ as its elements (def-coset, def-index), with product $(gN)(hN):=ghN\mathrm{.}$ 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 $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$).

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

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $\mathrm{\mid }G\times H\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }$. (For finite groups $G$ and $H$, $\mathrm{\mid }G\times H\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }\mathrm{\mid }H\mathrm{\mid }$).

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 $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}=(\mathbb{Z}\mathrm{/}n,+)$ 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 $(\mathbb{Z}\mathrm{/}n,+)$ is the quotient group $(\mathbb{Z},+)\mathrm{/}n\mathbb{Z}$).