Problem: Let $G$ be a finite nonabelian group. Assume any two elements $x,y \in G$ conjugate to each other also commute, show that $G$ is not simple.
I write down a proof of this problem based on its Sylow subgroups, but not sure if it is right. Can anyone verify it for me? Thanks in advance:
Proof: If $G$ is a $p$-group, then we are done, since nonabelian $p$-groups are not simple.
If $G$ is not a $p$-group, write $|G| = p^em,\, p \nmid m$. Given any two Sylow $p$-subgroups $P$ and $Q$ of $G$, they are conjugate, hence commute with each other. Consider the group $S = \langle PQ \rangle$ generated by $P$ and $Q$. It is not difficult to show $S = \{xy\mid x \in P, y \in Q\}$, since $P$ and $Q$ commutes. Also, $S$ is a $p$-group containing both $P$ and $Q$, because $(xy)^{p^e} = x^{p^e}y^{p^e} = 1$. By the maximality of Sylow subgroups, $S = P$ and $S = Q$. Hence $P = Q$, if follows $P \lhd G$, and $P \neq G$ for $|P| \lt |G|$. That is, $G$ is not simple.
edit: I try to correct my former proof (maybe I should say write down a new proof, but I think there is some connection between these two) and it seems to suffice now.
Corrected Proof: Given a conjugate class $x^G$ of $G$, consider the subgroup $H$ generated by its elements. Since the generators of $H$ commutes with each other, $H$ is abelian, so $H \neq G$, otherwise contradicting $G$ being nonabelian. Thus, $H$ is a proper subgroup of $G$.
For any element $h \in H$, and write $h = x_1x_2\dots x_m$ where $x_i \in x^G, i=1,2,\dots,m$. Then for any $g\in G$, $g^{-1}hg = (g^{-1}x_1g)(g^{-1}x_2g)\dots (g^{-1}x_mg)$ is also a product of elements in $x^G$, and the normality of $H$ follows.
We have $y^g=x$, then $xy=yx$. Based on your proof, it is not clear, why should $S=\langle P,Q\rangle$ be a $p$-group. Because we can not say that $x\in C_G(Q)$ for every $x\in P$. All we know that $x$ commutes with $x^g\in Q$.