A p-primary component has the full p-power order and is the unique subgroup of that order

Given: The objects and hypotheses in the statement.

Let $G$ be an abelian group and $p$ a prime. Its $p$-primary component is $G(p)=\{g\in G:g^{p^k}=e\text{ for some }k\in \mathbb{N}\}\mathrm{.}$ Thus the identity is included by $k=0$. In additive notation, $G(p)=\{g:p^{k}g=0\text{ for some }k\in \mathbb{N}\}$. Powers and element orders use def-group-power and def-order-in-a-group. No finiteness or maximality is part of the definition. (The p-primary component of an abelian group).

Let $G$ be a finite abelian group and let $p$ be a prime dividing $\mathrm{\mid }G\mathrm{\mid }$. Then $G$ contains an element, and hence a subgroup, of order $p$. (Cauchy's theorem for finite abelian groups).

Let $G$ be a finite group and $H\le G$. Then $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }\mathrm{.}$ Consequently, under the canonical embedding $\iota :\mathbb{N}\to \mathbb{Z}$, $\mathrm{\mid }H\mathrm{\mid }$ divides $\mathrm{\mid }G\mathrm{\mid }$. (Lagrange's theorem: $\mathrm{\mid }G\mathrm{\mid }=[G:H]\mathrm{\mid }H\mathrm{\mid }$ for every subgroup $H$ of a finite group $G$).

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 $\mathrm{\mid }G\mathrm{\mid }\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$).

Correspondence theorem: subgroups of $G\mathrm{/}N$ correspond to subgroups of $G$ containing $N$, with normality preserved. For $N⊴G$, the maps $H\mapsto H\mathrm{/}N$ and $K\mapsto {\pi }^{-1}(K)$ are inverse inclusion-preserving bijections between subgroups $H$ with $N\le H\le G$ and subgroups $K\le G\mathrm{/}N$; they preserve normality. (Correspondence theorem: subgroups of $G\mathrm{/}N$ correspond to subgroups of $G$ containing $N$, with normality preserved).

Second isomorphism theorem for groups: $H\mathrm{/}(H\cap N)\cong HN\mathrm{/}N$. If $H\le G$ and $N⊴G$, then $H\mathrm{/}(H\cap N)\cong HN\mathrm{/}N\mathrm{.}$ (Second isomorphism theorem for groups: $H\mathrm{/}(H\cap N)\cong HN\mathrm{/}N$).

If $G$ is abelian and $N⊴G$, then $G\mathrm{/}N$ is abelian. (Every quotient group of an abelian group is abelian).

Let $G$ be a group (def-group) with identity $e$, let $g,h\in G$, and let powers be as in def-group-power. For all $m,n\in \mathbb{Z}$: 1. $g^{m+n}=g^m{g}^n$; 2. $g^{-m}=(g^m{)}^{-1}$; 3. $(g^m{)}^n=g^{mn}$; 4. $g^m{g}^n=g^n{g}^m$: any two powers of one element commute; 5. if $gh=hg$ then $(gh{)}^n=g^n{h}^n$. Claim 5 is false in general without its hypothesis: in a group in which $g$ and $h$ do not commute the equation can fail already at $n=2$, and a witness is recorded on the companion page. Claims 1 and 3 hold in any monoid (def-semigroup-and-monoid) for exponents in $\mathbb{N}$, and so does claim 5 for exponents in $\mathbb{N}$ under the same commuting hypothesis; only the extension to negative exponents needs inverses. (Exponent laws in a group: $g^{m+n}=g^m{g}^n$ and $(g^m{)}^n=g^{mn}$ for all $m,n\in \mathbb{Z}$, and $(gh{)}^n=g^n{h}^n$ when $g$ and $h$ commute).