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Theorem: If $G$ is a finite group, then $c_a=o(G)/o(N(a))$; in other words, the number of elements conjugate to $a$ in $G$ is the index of the normalizer of $a$ in $G$.

Corollary: $o(G)=\sum \limits \dfrac{o(G)}{o(N(a))}$ where this sum runs over one element $a$ in each conjugate class.

Problem: Using Theorem as a tool, prove that if $o(G)=p^n$, $p$ a prime number, then $G$ has a subgroup of order $p^{\alpha}$ for all $0\leq \alpha \leq n$.

Can anyone help how to solve that problem? I was trying to use that center of $p$-group is nontrivial and induction of $n$ and tried to mix that with the class equation, but was not able to complete it.

Would be very thankful if somebody will demonstrate the solution of that problem.

RFZ
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  • Hi @RFZ! I think the answers given above are quite instructive here. – Arnaud Mortier Feb 09 '18 at 18:33
  • @ArnaudMortier, I guess that answer there uses Sylow's theorem. I would like to solve it via the class equation but I am not able to do it. – RFZ Feb 09 '18 at 18:35
  • I don't think so, the key ingredient is the class equation, see in step $3$ (in the answer that was selected). – Arnaud Mortier Feb 09 '18 at 21:53
  • @ArnaudMortier, Dear Arnaud! Could you help with question, please? https://math.stackexchange.com/questions/2652102/in-p-groups-subgroup-of-order-1-less-are-always-normal/2652108?noredirect=1#comment5477504_2652108. Namely why the first case never occurs? – RFZ Feb 15 '18 at 20:11
  • Hi! Lord Shark has removed his claim that the first case never occurs. Does his answer make sense to you now? – Arnaud Mortier Feb 15 '18 at 20:28
  • @ArnaudMortier, Hi! Not at all. I've got that $|\psi(H)|=1,p$. Why we rule out the case $|\psi(H)|=1$? After that we have that $|\psi(H)|=p$ then $\psi(H)$ is cyclic. How it follows that $H$ is normal in $G$? – RFZ Feb 15 '18 at 20:40
  • In the end the reason why $ H $ is normal is because it is the kernel of something. – Arnaud Mortier Feb 15 '18 at 22:38
  • The case "1" is excluded because an element not in $ H $ cannot have a trivial image. – Arnaud Mortier Feb 15 '18 at 22:44
  • In the case "p" the isomorphism $ G/ker \Psi \simeq Im \Psi $ tells you the order of the kernel, and since $ H $ contains the kernel and has the same sizel, they are equal. – Arnaud Mortier Feb 15 '18 at 22:48
  • @ArnaudMortier, Dear Arnaud! I have some difficulties with case "p". In this case we have, $o(\psi(H))=p$ and since we have mapping $\psi: H\to S_p$ then we have $H/ \text{Ker}\psi \cong \text{Im}\psi$ so $o(H)=o(\text{Ker}\psi)o(\text{Im}\psi)$ then $o(\text{Ker}\psi)=p^{n-2}$, right?

    We see that $H$ and $\text{Ker}\psi$ have different sizes. But you said that $H$ and kernel have identical size. What's wrong with my reasoning? I think that it is true.

     – RFZ Feb 17 '18 at 15:26
  • The domain of $\Psi$ is $G$, not $H$. – Arnaud Mortier Feb 17 '18 at 15:54
  • @ArnaudMortier, Yes but I do not understand the relation of your remark to my question? What's wrong? I have considered the mapping $\psi$ restricted to $H$. I do not see nothing wrong with that. Could you explain it in more detailed form, please? It's quite difficult for me since I am pondering on that problem the third day. – RFZ Feb 17 '18 at 16:04
  • Why would you restrict $\Psi$ to $H$? Note that if you do that, you change the image as well! The image becomes trivial, in fact. – Arnaud Mortier Feb 17 '18 at 16:16
  • @ArnaudMortier, Yes, indeed I am consider mapping $\Psi$ restricted to $H$? I do not see nothing bad in that. In this case the image will be just $\Psi(H)$, right? I have shown that $o(\Psi(H))=1 \ \text{or} \ p$. I just cannot understand what's wrong with my reasoning. Could you clarify it, please? – RFZ Feb 17 '18 at 16:21
  • Restricting $\Psi$ to $H$ is not "bad" in itself, but it tells you nothing (you obtain the trivial map, and no additional information on $H$). The unrestricted $\Psi$ is not trivial, that is a crucial information. – Arnaud Mortier Feb 17 '18 at 16:27
  • I added an answer, everything should be clear from there. – Arnaud Mortier Feb 17 '18 at 16:39
  • @ArnaudMortier, Thanks for answer! I have got it finally. Thanks a lot for that. It was a great help for me from your side :) – RFZ Feb 17 '18 at 18:19
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    You're very welcome :) it's good practice for me as well to try and make things clear. – Arnaud Mortier Feb 17 '18 at 23:49

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