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This question already has an answer here:

Let $G$ be a finite group and $p$ a prime number that divides the order of $G$. Let $h$ be the number of subgroups of $G$ of order $p$. Prove that there are $h(p-1)$ elements of order $p$ in $G$. Furthermore, prove $h \equiv 1 \pmod p$.

How should I solve the second part? This is what I did to prove the first:

Let $H$ be a subgroup of $G$ of order $p$. Because the order of an element in $H$ must divide the order of the group, its order must be $1$ or $p$. Since only the identity element has order $1$, the remaining $p-1$ elements have order $p$.

Assume that an element $x$ of order $p$ is contained in two subgroups $H_1$ and $H_2$. Then $x$ is a generator of both $H_1$ and $H_2$, so $\langle x \rangle = H_1 = H_2$, so an element of order $p$ is in only one such group.

So there are at least $h(p-1)$ elements of order $p$ in $G$. Assume there is an element $z$ of order $p$ which is not contained in any of the $h$ subgroups of order $p$. But we know that $\langle x \rangle$ is a subgroup of $G$ of order $p$. So all elements of order $p$ must be in one of those $h$ subgroups. Therefor, there are exactly $h(p-1)$ elements of order $p$ in $G$.

Now, how do I prove that $h \equiv 1 \pmod p$? This question is in a chapter about group actions, so that should maybe help. I'm assuming that I have to use the previously proven fact. And something that has to do with group actions. However, I'm not quite used to that concept and when it is useful to apply (more importantly, how to apply it). Could someone also explain to me its importance (especially in the case of this question)?

Thanks in advance.

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marked as duplicate by Martin Brandenburg, 23rd, Tom Oldfield, Douglas S. Stones, Micah May 9 '13 at 20:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Reason similar to the reasoning used to prove the sylow theorems. $\endgroup$ – davidlowryduda May 9 '13 at 19:44