Finite abelian $p$-group with only one subgroup size $p$ is cyclic

My goal is to prove this:

If $G$ is a finite abelian $p$-group with a unique subgroup of size $p$, then $G$ is cyclic.

I tried to prove this by induction on $n$, where $|G| = p^n$ but was not able to get very far with it at all (look at the edit history of this post to see the dead ends). Does anyone have any ideas for a reasonably elementary proof of this theorem?

Solution 1:

Assume $G$ is a finite abelian $p$-group with a unique subgroup of order $p$. I claim that if $a,b\in G$, then either $a\in\langle b\rangle$ or $b\in\langle a\rangle$. This will show that $G$ is cyclic, by considering all elements of $G$ one at a time.

Let $a,b\in G$; by exchanging $a$ and $b$ if necessary, we may assume that $|a|\leq |b|$, and we aim to prove that $a\in\langle b\rangle$. If $|a|\leq p$, then $\langle a\rangle$ is contained in the unique subgroup of order $p$, and hence is contained in $\langle b\rangle$, and we are done. So say $|a|=p^k$, $|b|=p^{\ell}$, $1\lt k\leq \ell$.

Let $t$ be the smallest nonnegative integer such that $a^{p^t}\in\langle b\rangle$. Note that since $a^{p^{k-1}}$ and $b^{p^{\ell-1}}$ are both of order $p$, the fact that $G$ has a unique subgroup of order $p$ means that $\langle a^{p^{k-1}}\rangle = \langle b^{p^{\ell-1}}\rangle$, so $t\leq k-1$; that is, $a^{p^t}\neq 1$. And since $a^{p^{t}}$ is of order $p^{k-t}$, we must have $\langle b^{p^{\ell-k+t}}\rangle = \langle a^{p^{t}}\rangle$. Let $u$ be such that $a^{p^{t}} = b^{up^{\ell-k+t}}$.

Now consider $x=ab^{-up^{\ell-k}}$. Note that since $k\leq \ell$, this makes sense. What is the order of $x$? If $x^{p^r}=1$, then $a^{p^r} = b^{up^{\ell-k+r}}\in\langle b\rangle$, so $r\geq t$ by the minimality of $t$. And $$x^{p^t} = a^{p^t}b^{-up^{\ell -k + t}} = b^{up^{\ell-k+t}}b^{-up^{\ell-k+t}} = 1.$$ So $x$ is of order $p^t$.

If $t\gt 0$, then $x^{p^{t-1}}$ has order $p$, so $x^{p^{t-1}}\in \langle b\rangle$. But $$x^{p^{t-1}} = a^{p^{t-1}}b^{-up^{\ell-k+t-1}},$$ so the fact this lies in $\langle b\rangle$ means that $a^{p^{t-1}}\in\langle b\rangle$. The minimality of $t$ makes this impossible.

Therefore, $t=0$, which means $x=1$. Thus, $a^{p^0} = a\in\langle b\rangle$, as desired.

Added. The above argument shows that an abelian group $A$ in which every element has order a power of $p$ and that contains a unique subgroup of order $p$ is locally cyclic; that is, any finitely generated subgroup of $A$ is cyclic. This includes some groups that are not finite or finitely generated, e.g. the Prüfer $p$-groups.

Solution 2:

A nice characterization of finite cyclic groups $G$ is:

Given any integer $m$, the number of solutions to $x^m=1$ in $G$ is at most $m$.

The proof is easy, and is mostly counting. If your group $G$ was non-cyclic, it would fail the above, so there is some $m$ with $G$ having more than $m$ solutions to $x^m=1$; let's call these solutions $\lbrace x_1,\ldots,x_k\rbrace$. Let's also make sure we pick the smallest $m$ that works. Of course, $G$ is a p-group, so $m$ is a multiple of $p$, say $m=np$. Can you show the collection $\lbrace x_1^n,\ldots,x_k^n\rbrace$ has more than $p$ distinct elements? And do you see how that contradicts your hypothesis?