Elements in non-Abelian simple Groups.

Question

Suppose $K$ is non-abelian and simple. Let $k\in K-\{e\}$. Then there exists $r\in K$ s.t. $rkr^{-1}\neq k$.

I'm not sure whether the following proof works:

Suppose by contradiction that for all $r\in K$, $rkr^{-1}=k$. This implies that the cyclic group generated by $k$, i.e., $\langle k \rangle ,$ is normal, i.e., $rk^mr^{-1}=k^m$. Hence, $\langle k\rangle={1}$ or $\langle k \rangle = K$. Since $k\neq 1$, this implies $\langle k\rangle=K$. Hence contradiction.

Is this valid?

Answer 1

This looks good to me, though I might emphasize at the end that the contradictory conclusion is that $K$ is cyclic and hence abelian.

Here's another argument that proceeds contrapositively, that is, again assumes there is no such $r$.

Hint Rearranging gives that $rk = kr$ for all $r \in K$, that is, $k$ is in the center $Z(K)$ of $K$, and in particular $Z(K)$ is nontrivial.

On the other hand, since $K$ is nonabelian, $Z(K) \lneq K$. The center of any group is a normal subgroup, so $Z(K)$ is a normal subgroup of $K$ that is neither $\{1\}$ nor $K$, and hence $K$ is not simple.