A maximal-order cyclic subgroup splits off a finite abelian p-group

Given: The objects and hypotheses in the statement.

Let $G$ be a nontrivial finite abelian $p$-group. If $G$ has exactly one subgroup of order $p$, then $G$ is cyclic. (A nontrivial finite abelian p-group with a unique subgroup of order p is cyclic).

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 $P$ be a property of naturals such that for every $n \in \mathbb{N}$, if $P(m)$ holds for all $m < n$ then $P(n)$. Then $P(n)$ holds for all $n \in \mathbb{N}$. (At $n = 0$ the hypothesis is vacuous, so $P(0)$ is forced.) (Strong (complete) induction).

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 $(gN)(hN)=ghN$).

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).

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 $|G|/|N|$).

Let $G$ be a group and $g \in G$, with integer powers as in def-group-power. Then $\langle g \rangle \;=\; \{\, g^{n} \;:\; n \in \mathbb{Z} \,\} ,$ the cyclic subgroup generated by $g$ (def-generated-subgroup) being exactly the set of integer powers of $g$. Consequently every cyclic group is abelian, and so is every cyclic subgroup of any group. ($\langle g \rangle = \{\, g^{n} : n \in \mathbb{Z} \,\}$, and every cyclic group is abelian).

Let $G$ be a group and let $N_0,\ldots,N_{r-1}$ be normal subgroups, where $r\in\mathbb N$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap\langle N_j:j<r,\ j\ne i\rangle=\{e\}.$ The empty family is an internal direct product of the trivial group. For two subgroups of an abelian group this says $G=HK$ and $H\cap K=\{e\}$; in additive notation one writes $G=H\oplus K$. Normal subgroups and generated subgroups are those of def-normal-subgroup and def-generated-subgroup, and the comparison product is def-external-direct-product-of-groups. (Internal direct products of finitely many normal subgroups).

Let $G$ be a group, $g \in G$, and let orders be as in def-order-in-a-group. Throughout, a natural number written where an integer is expected means its image under the embedding $\iota : \mathbb{N} \to \mathbb{Z}$ of lem-nat-embeds-int. Finite order. Suppose $\operatorname{ord}(g) = n$ with $n \in \mathbb{N}$, $n \ge 1$. Then: 1. for every $k \in \mathbb{Z}$, $g^{k} = e$ if and only if $k = qn$ for some $q \in \mathbb{Z}$, that is, if and only if $n \mid k$ (thm-division-algorithm-in-z); 2. the powers $g^{0}, g^{1}, \dots, g^{n-1}$ are pairwise distinct: if $i, j \in \mathbb{N}$ with $i < n$, $j < n$ and $g^{i} = g^{j}$, then $i = j$; 3. $\langle g \rangle = \{\, g^{s} : s \in \mathbb{N},\ s < n \,\}$ and $\langle g \rangle \approx n$; so $\langle g \rangle$ is finite with $|\langle g \rangle| = n = \operatorname{ord}(g)$. Infinite order. If $\operatorname{ord}(g) = \infty$ then for $j, k \in \mathbb{Z}$, $g^{j} = g^{k}$ implies $j = k$; so the integer powers of $g$ are pairwise distinct and $\langle g \rangle$ is not finite. (If $\operatorname{ord}(g) = n$ then $g^{k} = e$ iff $k$ is an integer multiple of $n$, the powers $g^{0}, \dots, g^{n-1}$ are distinct, and $\langle g \rangle$ has exactly $n$ elements; if $g$ has infinite order then $g^{j} = g^{k}$ only for $j = k$).