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Question. Let $G$ be a finite group and $p$ be the smallest prime dividing $|G|$.

Let $x$ be n element of order $p$ in $G$. Assume that there exists an element $h\in G$ such that $hxh^{-1}=x^{10}$.

Show that $p=3$.


What I have done so far is use the fact that $o(hxh^{-1})=o(x)=p$ and $o(x^{10})=p/\gcd(p,10)$.

By hypothesis we have $p=p/\gcd(p,10)$, giving $\gcd(p,10)=1$.

Here I am stuck.

I have not been able to make use of the fact that $p$ is the smallest prime divisor of $|G|$.

Can somebody help?

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1 Answer 1

up vote 6 down vote accepted

Let $n$ be the order of $h$. Then $$x = h^n x h^{-n} = x^{10^n}$$

So $10^n \equiv 1 \mod{p}$. Now apply the fact that $\gcd(p-1, n) = 1$.

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Oh nice! That's how you used the fact that $p$ is the smallest prime divisor of $|G|$. –  caffeinemachine Mar 2 at 10:36

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