A nontrivial finite abelian p-group with a unique subgroup of order p is cyclic

Given: The objects and hypotheses in the statement.

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

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

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, $g\in G$, 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 $j=k$).