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 $G(p)=\{g\in G:g^{p^k}=e\text{ for some }k\in\mathbb N\}.$ Thus the identity is included by $k=0$. In additive notation, $G(p)=\{g:p^kg=0\text{ for some }k\in\mathbb N\}$. 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 $|G|$. 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 $|G|=[G:H]\,|H|.$ Consequently, under the canonical embedding $\iota:\mathbb N\to\mathbb Z$, $|H|$ divides $|G|$. (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$).

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 $|G|/|N|$).

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

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

If $G$ is abelian and $N\mathrel{\trianglelefteq}G$, then $G/N$ 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).