# Can we conclude that this group is cyclic? [duplicate]

Let $G$ be a finite group. If, for each positive integer $m$, the number of solutions of the equation $x^m = e$ in $G$, where $e$ is the identity element, is at most $m$, then can we conclude that $G$ is cyclic?

## marked as duplicate by Mikko Korhonen, Stefan Hansen, Julian Kuelshammer, dfeuer, azimutOct 7 '13 at 6:57

• For abelian groups this follows immediately from the structure theorem. – N. S. Oct 6 '13 at 22:31
• N.S., can't we give a proof independent of the structure theorem? – Saaqib Mahmood Oct 6 '13 at 22:44
• YACP, how? Can you please elaborate? – Saaqib Mahmood Oct 6 '13 at 22:46
• Where did this question come from? – dfeuer Oct 6 '13 at 22:46
• Check the notes at the end of my answer, I forgot too many things about Syllow Theorems to give a proof, but I think the last claim is true, which would prove the result in the general settings. – N. S. Oct 6 '13 at 22:47

Let $|G| = n$ and $m \mid n$. Define $G_m = \{g \in G : |g| = m\}$.

Suppose $G_m$ is not empty and let $g \in G_m$. By assumption, $x^m = e$ has at most $m$ solutions in $G$. Since each element of $\langle g \rangle$ is a solution, there cannot be more. From those, there are $\varphi(m)$ elements with order exactly $m$. This is a basic result on cyclic groups, where $\varphi$ is Euler's totient function.

It follows that $|G_m| = \varphi(m)$ if $G_m$ is not empty. If $G_m$ is empty, then $|G_m| = 0$. In general, $|G_m| \le \varphi(m)$.

Now we have: $$n = |G| = \sum_{m \mid n} |G_m| \le \sum_{m \mid n}\varphi(m) = n$$

Therefore, the inequality is in fact an equality. This shows that $|G_n| = \varphi(n) > 0$. Hence $G$ is cyclic since it contains an element of order $n$.

• @SaaqibMahmuud The inequality being equality shows that $|G_n| =\phi(n) >0$. – N. S. Oct 7 '13 at 4:50
• @YoavFridman The very first line. We only consider elements with order $m \mid n$. – Ayman Hourieh Dec 19 '13 at 17:31
• @YoavFridman Lagrange's theorem says that if the order of a group is $n$, then the order of every element in this group divides $n$. This is why the union of all sets $G_m$ is equal to $G$ in my answer. – Ayman Hourieh Dec 19 '13 at 23:25
• @YoavFridman Since $g$ has order $m$, each element $x \in \langle g\rangle$ has order at most $m$. Hence $x^m = e$. – Ayman Hourieh Dec 21 '13 at 10:50
• @YoavFridman It's because every member of $G$ must be in one and only one $G_m$. Think about the definition of $G_m$. – Ayman Hourieh Dec 21 '13 at 11:51

Short note. If $G$ is a $p$-group, the result is true. Indeed, if $|G|=p^k$, then if $G$ is not cyclic, it follows that every element in $G$ we have

$$x^{p^{k-1}}=e \,.$$

Thus this contradicts the statement with $m=p^{k-1}$.

Now, for abelian groups, by the structure Theorem, the group is a product of $p$ groups. Use the above result for each $p$ groups, and you are done.

In the non-abelian case, the above result shows that all the $p$-Syllow subgroups are cyclic. Moreover, $G$ can only have one $p$-syllow subgroup, otherwise you get more than $p^k$ solutions to $x^{p^k}=e$.

So the problem reduces to the following simple problem (which should be obvious, but it is not to me). If for every $p$, $G$ has an unique $p$ Syllow subgroup which is cyclic, is $G$ (abelian) cyclic?

• N.S., thank you, but I wonder if we could prove this assertion through more elementary means. – Saaqib Mahmood Oct 6 '13 at 22:47
• @SaaqibMahmuud What do you mean by elementary ? No group Theory? :P – N. S. Oct 6 '13 at 22:48
• N.S., by elementary I mean not involving notions such as Sylow groups etc. – Saaqib Mahmood Oct 7 '13 at 4:47