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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?

Saaqib Mahmood
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2 Answers2

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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 $n$.

Ayman Hourieh
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  • Ayman Hourieh, so far so good, but how do we conclude from this last equality that $G$ necessarily has an element of order $n$? – Saaqib Mahmood Oct 07 '13 at 04:49
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    @SaaqibMahmuud The inequality being equality shows that $|G_n| =\phi(n) >0$. – N. S. Oct 07 '13 at 04:50
  • Where you are using here Lagrange theorem? (I know that at this proof we need to use Lagrange Theorem...) – CS1 Dec 19 '13 at 16:48
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    @YoavFridman The very first line. We only consider elements with order $m \mid n$. – Ayman Hourieh Dec 19 '13 at 17:31
  • @AymanHourieh - Why, how can you assume that $m\mid n$? (I got this Q as H.W. and it's doesn't mention there, BUT, what is mention there - is that I need to use Lagrange theorem, so what I'm asking you, is how do you use it at very first line [as you mention]). Thank you! – CS1 Dec 19 '13 at 20:48
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    @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
  • And how do you know that $\sum_{m \mid n}\varphi(m) = n$? Thank you! – CS1 Dec 20 '13 at 09:32
  • @YoavFridman It's a well-known formula. Here is a proof that uses group theory. – Ayman Hourieh Dec 20 '13 at 11:05
  • @AymanHourieh - Oh, I understand. thank you!! I didn't understand that you base on Euler's totient function... – CS1 Dec 20 '13 at 15:35
  • @AymanHourieh - one more Q, how do you know that: Since each element of $$ is a solution..... I'm asking about the $$, you assume it's cyclic? right? if yes, how you can assume it? Thank you! – CS1 Dec 21 '13 at 07:36
  • @AymanHourieh - And why each element of $\langle g \rangle$ is a solution? I don't understand this too... – CS1 Dec 21 '13 at 07:42
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    @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
  • @AymanHourieh - One more thing that I don't understand: $|G| = \sum_{m \mid n} |G_m|$ how you conclude this? – CS1 Dec 21 '13 at 11:40
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    @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
  • @AymanHourieh - So, if I'm understanding the proof right, $\langle g \rangle=G$? – CS1 Dec 21 '13 at 11:54
  • Or I miss something... – CS1 Dec 21 '13 at 11:54
  • @AymanHourieh - There is something that I still don't understand: if $\langle g \rangle\ne G$ so how your proof proves that $G$ is cyclic?? If you can clear this point for me, it will be great, thank you... – CS1 Dec 21 '13 at 12:31
  • @YoavFridman I edited my answer. See the last sentence. – Ayman Hourieh Dec 21 '13 at 12:53
  • @AymanHourieh - Thank you for your edit, but I still don't understand few things:
    1. Why it's means that $G$ is cyclic?
    2. How do you jump from $|G_m|$ to $|G_n|$?
    3. This - $\sum_{m \mid n} |G_m| \le \sum_{m \mid n}\varphi(m)$ how do you get this?

    Thank you and sorry for all of my Q...

     – CS1 Dec 21 '13 at 14:31
  • @AymanHourieh, I'm sorry that I'm asking you a lot of Q, but why you proof proves that $G$ is cyclic? I don't understand this... I hope that if I'll understand this, the proof will be much more clear to me... – CS1 Dec 21 '13 at 16:06
  • @AymanHourieh, do you saw my questions? I'm sorry if it's too much, but I really don't understand few things at the proof... Please answer me... – CS1 Dec 21 '13 at 20:53
  • @YoavFridman I'm sorry but I can't respond in length at the moment. I'll try later today. – Ayman Hourieh Dec 21 '13 at 20:54
  • @AymanHourieh Thank you so much!! I'm waiting for your answers! – CS1 Dec 21 '13 at 21:29
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    @YoavFridman A group of order $n$ is cyclic if it contains an element of order $n$. This is how the proof is concluded. To be honest, I don't have anything more to say apart from what's already been said here and on your other question. I suggest you review cyclic groups before coming back to this question. Once done, try to read each step in the proof and justify it yourself. I feel that you need to review and fill in some gaps that are making this proof more difficult than it actually is. – Ayman Hourieh Dec 22 '13 at 15:51
  • @AymanHourieh, you helped me a lot!!! Thank you so much!!! now I'm understanding it!!! – CS1 Dec 24 '13 at 10:09
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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.
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