Understanding a proof about finite $p$-groups

Question

I can't follow the reasoning of the author,in this proof:

let $G$ be a finite $p-$group. If $H$ is a proper subgroup of G, then $H<N_G(H)$

(clearly $N_G(H)$ is the normalizer of $H$ and p is a prime)

Proof:

if $H$ is normal then $N_G(H)=G$ and the theorem is true. If $X$ is the set of all the conjugates of $H$, then we may assume that $|X|=[G:N_G(H)] \neq 1$

and up to this I'm ok

Now $G$ acts on $X$ by conjugation and, since $G$ is a p-group, every orbit of X ha size a power of p. As $\lbrace H \rbrace$ is a orbit of size 1, there are at least one conjugate $H^g \neq H$ with $\lbrace H^g \rbrace$ also an orbit of size 1.

And this part gives me some problems. Why the orbit of H has size 1? It would be normal in this case right? Or it is a typo and the group acting on X is H?

Now $(H^g)^a=H^g$ for every $a \in H$, and so $a^g \in N_G(H)$ for,all $a \in H$. But $H^g \neq H$ gives at least one $a \in H$ with $ a^g \notin H$, and so $H < N_G(H)$.

I can't understand where this $a$ come from, is a result of the fact that by absurd if it is would be true, then $g \in N_g(H)$ and so the two would be the same orbit?

if someone could clarify this proof I would appreciate :)

Answer 1

Now $G$ acts on $X$ by conjugation and, since $G$ is a $p$-group, every orbit of $X$ has size a power of $p$. As $\{H\}$ is an orbit of size $1$, there is at least one conjugate $H^g\neq H$ with $\{H^g\}$ also an orbit of size $1$.

This part contains a typo or mistake. The group we want to act on $X$ by conjugation is $H$. Then it is clear that $\{H\}$ is a singleton orbit, and since the cardinality of $X$ is a multiple of $p$, there must be other singleton orbits (at least $p-1$). We fix one of them.

Now $(H^g)^a=H^g$ for every $a\in H$, and so $a^g\in N_G(H)$ for,all $a\in H$. But $H^g\neq H$ gives at least one $a\in H$ with $a^g\notin H$, and so $H < N_G(H)$.

$H^g$ being a singleton orbit means $(H^g)^a = H^g$ for all $a \in H$, and hence

$$H^{gag^{-1}} = \bigl((H^g)^a\bigr)^{g^{-1}} = (H^g)^{g^{-1}} = H$$

for all $a \in H$, i.e. $gag^{-1} \in N_G(H)$ for all $a \in H$. But $\{ gag^{-1} : a \in H\} = H^{g^{-1}}$, so

$$(H \cup H^{g^{-1}}) \subset N_G(H).$$

By the choice of $H^g$ as a conjugate of $H$ different from $H$ that is fixed by the action, we then have $H = (H^g)^{g^{-1}} \neq H^{g^{-1}}$, and therefore $H \subsetneq N_G(H)$.