A property in finite $p$-groups

Question

I study a paper about finite $p$-groups and I could not understand the following property which is used in it. Any comment or answer will be appreciated!

Let $G$ be a finite non-abelian $p$-group and $x\in G$. If $g\in N_G(\langle x\rangle)$, then $[x,g]\in \langle x^p\rangle$.

( Note that $N_G(\langle x\rangle)=\{g\in G | g^{-1}\langle x\rangle g=\langle x\rangle\}$ and $[x,g]=x^{-1}g^{-1}xg$).

Many thank's.

Answer 1

I offer a slightly different proof of this fact.

Let $G$ be a $p$-group. Let $x,g\in G$ and $g^{-1}xg=x^\alpha$, $\alpha$ is a positive integer not divisible by $p$.

If the order of $g$ is $p^k$, then $$ x=g^{-p^k}xg^{p^k}=x^{\alpha^{p^k}}. $$ It follows that $\alpha^{p^k}\equiv1\pmod p$. On the other hand, due to Fermat's little theorem, we have $\alpha^{p^k}\equiv\alpha\pmod p$. Therefore, $\alpha\equiv1\pmod p$ and hence. $$ [x,g]=x^{-1}\cdot g^{-1}xg=x^{-1}x^\alpha=x^{\alpha-1}\in\langle x^p\rangle. $$

Answer 2

If $x=e$, there is nothing to do. Assume $x\neq e$.

Let $g\in N_G(\langle x\rangle)$.

Note that since $[x,g]\in\langle x\rangle$, then $x$ commutes with $[x,g]$. Since $$[ab,c] = [a,c]^b[b,c]$$ holds in any group, it follows that $[x^n,g]=[x,g]^n$ for all integers $n$. Indeed, we have that $$[x^{k+1},g] = [xx^k,g] = [x,g]^{x^k}[x^k,g] = [x,g][x^k,g],$$ so if $[x^k,g]=[x,g]^k$, then $[x^{k+1},g]=[x,g]^{k+1}$. And $[x^{-1},g] = [g,x]^{x^{-1}}=[g,x]=[x,g]^{-1}$, so the result also follows for negative powers.

That means that the function $f\colon \langle x\rangle \to \langle x\rangle$ given by $f(x^a) = [x^a,g]$ is in fact a group homomorphism.

Let $H=\langle x,g\rangle$. Then $\langle x\rangle\triangleleft H$, and therefore $\langle x\rangle\cap Z(H)\neq\{e\}$, since any nontrivial normal subgroup of a $p$-group intersects the center nontrivially. (This is where I am using that $x\neq e$). Thus, the kernel of $f$ is nontrivial, which means that $f$ is not surjective.

Since $f$ is not surjective, its image is not all of $\langle x\rangle$, and therefore the image is contained in $\langle x^p\rangle$ (which is the unique maximal subgroup of $\langle x\rangle$) as desired.