Error in proving subgroup is normal

Question

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.

Answer 1

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.

Answer 2

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.