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

Let $P$ 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 $N⊴G$ 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 $(gN)(hN)=ghN$).

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).

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

Let $G$ be a group and $g\in G$, with integer powers as in def-group-power. Then $⟨g⟩=\{g^n:n\in \mathbb{Z}\},$ 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. ($⟨g⟩=\{g^n:n\in \mathbb{Z}\}$, and every cyclic group is abelian).

Let $G$ be a group and let $N_0,\dots ,N_{r-1}$ be normal subgroups, where $r\in \mathbb{N}$. They form an internal direct product when they generate $G$ and, for each $i<r$, $N_i\cap ⟨N_j:j<r,j\mathrm{\ne }i⟩=\{e\}\mathrm{.}$ 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 $\mathrm{ord}(g)=n$ 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. $⟨g⟩=\{g^s:s\in \mathbb{N},s<n\}$ and $⟨g⟩\approx n$; so $⟨g⟩$ is finite with $\mathrm{\mid }⟨g⟩\mathrm{\mid }=n=\mathrm{ord}(g)$. Infinite order. If $\mathrm{ord}(g)=\mathrm{\infty }$ then for $j,k\in \mathbb{Z}$, $g^j=g^k$ implies $j=k$; so the integer powers of $g$ are pairwise distinct and $⟨g⟩$ is not finite. (If $\mathrm{ord}(g)=n$ then $g^k=e$ iff $k$ is an integer multiple of $n$, the powers $g^0,\dots ,g^{n-1}$ are distinct, and $⟨g⟩$ has exactly $n$ elements; if $g$ has infinite order then $g^j=g^k$ only for $j=k$).