Prove that $3$ is the smallest prime dividing the order of the group.

Question

Let $G$ be a finite group where $p$ is the smallest prime dividing the order of the group. Now, let $x$ be an element of order $p$ in $G$ such that $hxh^{-1}=x^{10}$ for some $h \in G$. Then we need to show that $p=3$

My approach to the solution goes like this: let $H$ be a subgroup generated by $h$ and $K$ be the subgroup generated by $x$. Now we consider the group action of conjugation from,

$H \times K \to \text{Aut}(K)$.

We define a map $\phi$ such that,

$\phi:H \to \text{Aut}(K)$ where $\phi(h^k):x \mapsto h^k x h^{-k}$.

I am kind of stuck after this. I can understand that order of Aut($K$) is $p-1$.

I have also been able to show using some calculations that order of the element($p$) will be either $3$ or $11$.

However I am not sure of how to continue after this. This is a practice problem to revise automorphism of groups and I have absolutely forgotten all the theory. I need some help with the basic.

Answer 1

If you have already shown that the order of the element $x$ is either $3$ or $11$, then suppose that $ord_G(x)=11$, so $x^{11}=1$. You already know that the order of $G$ is odd. There is a theorem that when the order of a group is odd, then any nonidentity element $x$ is not conjugate to $x^{-1}$ (see e.g. here: An element of a group $G$ is not conjugate to its inverse if $\lvert G\rvert$ is odd) . By our assumption, we get $hx=x^{10}h$, so $xhx=x^{11}h=h$. It follows $x=hx^{-1}h^{-1}$. As the order of $x$ is $11$, the element $x$ certainly is not the identity. So our $x\neq 1$ is a conjugate to its inverse. Thus by our above theorem, we get a contradiction. So the order of $x$ must be $3$ by the work you've already completed.

Answer 2

Put $H=\langle x,h \rangle$ and let $N=\langle x \rangle$. Then $H$ is a subgroup of $G$ and $N \subseteq H$. Since $hxh^{-1}=x^{10}$, we see that $N \unlhd H$. Now, $N_H(N)/C_H(N)=H/C_H(N)$ embeds homomorphically into Aut$(N) \cong C_{p-1}$. Since $p$ is the smallest prime dividing the order of $G$, the same holds true for the subgroup $H$. Hence $H=C_H(N)$, that is $N \subseteq Z(H)$. It follows that $hxh^{-1}=x=x^{10}$, from which we get $x^9=1$, so $p=3$.