# Error in proving subgroup is normal

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.

• $|G/H|$ 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. – Randall Nov 28 '18 at 3:16
• "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. – астон вілла олоф мэллбэрг Nov 28 '18 at 3:21
• 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. – AdditIdent Nov 28 '18 at 5:41
• Now that I think about it, I don't see how this action is an action at all. – Randall Nov 28 '18 at 20:51

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.
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.