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Let $G$ be a $p$-group and let $H$ be a subgroup of $G$ of index $p$ ($p$ is prime). Prove that $H$ is normal.

I have tried to prove it, but accidentally have proven the opposite by some error:

Let $G$ act on $G/H$ by conjugation. We have proven as a lemma that the number of fixed points of this action is equivalent to $|G/H|$ mod $p$, because $G$ is a $p$-group (see here). Thus, since $|G/H| = p \equiv 0 $ mod $p$, it follows there are no fixed points. Thus $H$ is not a fixed point of the action, so there exists some $g\in G$ such that $g.H\neq H \implies gHg^{-1}\neq H$. So $H$ is not normal in $G$.

I have tried to find the error but cannot. Where have I gone wrong? Is the action itself not well-defined? I am not sure how to proceed.

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    $\begingroup$ $|G/H|$ is $p$, and the number of fixed points is congruent to this mod $p$. But this is satisfied by both $0$ as well as $p$: they're both congruent to $p$ mod $p$. So, you can't automatically say there are no fixed points. $\endgroup$ – Randall Nov 28 '18 at 3:16
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    $\begingroup$ "Since $|G/H| \equiv 0 \mod p$, there are no fixed points". No, it just means that the number of fixed points is multiple of $p$, like $2p,p, 0$ etc. $\endgroup$ – астон вілла олоф мэллбэрг Nov 28 '18 at 3:21
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    $\begingroup$ It doesn't even make sense to do quotient before knowing nH$ is normal, it is exactly what to prove. The statement indeed follows directly from Sylow Theorem. $\endgroup$ – AdditIdent Nov 28 '18 at 5:41
  • $\begingroup$ Now that I think about it, I don't see how this action is an action at all. $\endgroup$ – Randall Nov 28 '18 at 20:51
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Thinking more about this, there are two issues with your argument. The one raised already is that the orbit-stabilizer techniques show that the number of fixed points must be congruent to $0 \bmod p$, but this doesn't mean it has to be the integer $0$. Hence there could be fixed points.

I think the bigger issue is your action itself. If $H$ is not known to be normal (yet), then $G/H$ can only mean the set of left (or right, pick a side) cosets of $H$ in $G$. Standard actions of $G$ on this set in standard Sylow-ish proofs involve action by left multiplication, not conjugation. Indeed, if your action is "defined" by $$ g \cdot xH = g(xH)g^{-1} $$ this doesn't make any sense at all: how is $g(xH)g^{-1}$ again a left coset? There's no reason.

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  • $\begingroup$ You're correct, and it turns out it does work, but only because the conclusion that H is normal is true. One can prove that with a totally different argument, but I think an interesting thing about this one is it actually proves (once you know H is normal) that every left coset is fixed under conjugation. $\endgroup$ – MathTrain Nov 28 '18 at 21:08
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Randall commented correctly that the fact that $|G/H|=p$ does not mean that the number of fixed points is a multiple of $p$, not that it is zero.

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