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?

  • $\begingroup$ Have you prove that $G/H$ verifies the induction hypothesis? $\endgroup$ – Josué Tonelli-Cueto Nov 13 '11 at 9:59
  • $\begingroup$ Well $G/H$ has size $p^n$, which is finite and of a size the hypothesis applies to, and it is abelian because $G$ is assumed to be abelian. So it seems to satisfy the induction hypothesis. $\endgroup$ – Mani Nov 13 '11 at 10:03
  • $\begingroup$ One part of the induction hypothesis is that it has a unique subgroup of order $p$. $\endgroup$ – Josué Tonelli-Cueto Nov 13 '11 at 10:07
  • $\begingroup$ @Iasafro I've added an argument for that, see the edit above. $\endgroup$ – Mani Nov 13 '11 at 10:27
  • $\begingroup$ You have made a circular argument. You cannot conlude that $G/H$ has a unique subgroup of order $p$ by applying that it is cyclic, i.e. that has a unique generator $gH$, since $G/H$ being cyclic is a consecuence of the induction hypothesis, that is, of having $G/H$ a unique subgroup of order $p$. In brief, being cyclic has to be a consecuence of $G/H$ having a unique subgroup of order $p$ and not in the other sense. $\endgroup$ – Josué Tonelli-Cueto Nov 13 '11 at 10:37

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.

  • $\begingroup$ "And since $a^{p^{t}}$ is of order $p^{k-t}$, we must have $\langle b^{p^{\ell-k+t}}\rangle = \langle a^{p^{t}}\rangle$." Can you explain why that is true? $\endgroup$ – Stefan Dec 10 '11 at 12:54
  • $\begingroup$ @StefanWalter: Remember that $a^{p^t}\in\langle b\rangle$. It generates the unique subgroup of order $p^{k-t}$ (cyclic groups have unique subgroups of any given order dividing the size of the group). Now, $b$ is of order $p^{\ell}$, so $b^{p^{\ell-k+t}}$ also generates a subgroup of $\langle b\rangle$ of order $p^{k-t}$. Since there is only one such subgroup, $\langle b^{p^{\ell-k+t}}\rangle = \langle a^{p^t}\rangle$. $\endgroup$ – Arturo Magidin Dec 10 '11 at 20:57
  • $\begingroup$ Ok, thank you very much! $\endgroup$ – Stefan Dec 11 '11 at 0:18

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?


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.

  • 1
    $\begingroup$ If you use \langle and \rangle instead of < and >, then $\LaTeX$ gives the proper spacing; compare $\langle g_0\rangle$ with $<g_0>$. $\endgroup$ – Arturo Magidin Nov 13 '11 at 23:28
  • $\begingroup$ I am intrigued - why the additive notation? $\endgroup$ – user1729 Nov 14 '11 at 10:22
  • $\begingroup$ @ArturoMagidin: Thanks for the advice. $\endgroup$ – Josué Tonelli-Cueto Nov 14 '11 at 17:22
  • $\begingroup$ @user1729: Well, the group is abelian. Normally, in abelian cases I like to use additive notations. $\endgroup$ – Josué Tonelli-Cueto Nov 14 '11 at 17:23
  • $\begingroup$ Sorry, I didn't read that bit! I thought the group was non-abelian, and then additive notation would have been...odd... $\endgroup$ – user1729 Nov 14 '11 at 17:35

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