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

Question

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?

Answer 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.

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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.

Answer 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?

Answer 3

let's use induction on $n$. the base case is trivial. now if $|A|=p^{n}$ and B is the unique cyclic group, $A/B$ which is isomorphic to the image of $f:A\rightarrow A$ defined by $f(x)=px$. as a subgroup of $A$, $B$ is a unique subgroup of order p of $f(A)$ as well (it must have a subgroup of order p by it being a p-group, the rest follows from our hypothesis on $A$). by the induction hypothesis we conclude that $A/B$ is cyclic. suppose $<x+B>=A/B$. we can see that $<x>=A$,since $B$ is contained in $<x>$ (otherwise $B$ is not unique $p$ ordered group in $A$).

Answer 4

I will you give an answer that use more or less elementary techniques. (I will use additive notation.)

We will proceed by counter implication, i.e. showing that when $G$ is not cyclic then there are at least two subgroups of order $p$. Let $h$ be an element of $G$ which has maximum order $p^n$, and let $H=\langle h\rangle$. Now, when considering $$G/H$$ we have that since $G$ is not cyclic it is not trivial. So there is an element $g+H$ of order $p$ -obtaining this element is trivial since $G/H$ is an abelian $p$-group, given a non-identity element $g_0+H$ of order $p^r$ take $g+H=p^{r-1}(g_0+H$)-.

Now, we have that $$p(g+H)=H$$ i.e. that for some $a\in\mathbb{N}_0$ and $k$ coprime to $p$ $$pg=kp^a h$$ First, we have to note that $kh$ is also a generator of $H$ since $k$ is coprime to the order of $h$. So, we have that $$pg=p^a(kh)$$ has order $p^{n-a}$, which implies since $g$ is not $0_G$ that the order of $g$ is $p^{n-a+1}$. Thus, we must have $a\geq 1$ in order to not having $g$ grater order than $h$ in contradiction with the choice of $h$, which is one element of maximum order in $G$.

From here, we can take $$g_0=g+(p^{n-a}-k)p^{a-1}h\not\in H$$ since $g\not\in H$. And obtaining that $$pg_0=p(g+(p^{n-a}-k)p^{a-1}h)=pg+(p^{n-a}-k)p^{a}h=kp^a h+(p^{n-a}-k)p^{a}h=p^n h=0_G$$ which means that $g_0$ is an element of ordr $p$, i.e. that $$\langle g_0\rangle$$ is a subgroups of order $p$ distinct from $$\langle p^{n-1}h\rangle$$ as desired.

Finally, counterimplication gives the desired statement, i.e. that a finite abelian $p$-group with an unique subgroup of order $p$ is neccesarily cyclic.