Prove a group is not simple, and nilpotent I ran across this problem.  Suppose that $G$ is a group where any two elements that are conjugate commute with each other.  Then $G$ is not simple.  It goes on to state that, in fact, $G$ must be nilpotent.  I tried playing around with the Sylow theorem that says all $p$-sylows are conjugate, but couldn't work anything out.  Then I thought about showing that the commutator subgroup is proper, but again couldn't think of anything.
 A: The claim is not true as stated because the abelian simple groups $\mathbb Z/p\mathbb Z$ trivially have this property.
So assume $G$ is non-abelian and has the stated property. Let $1\ne a\in G$ and $H=\langle \,gag^{-1}\mid g\in G\,\rangle$. Then the generators commute, hence $H$ is abelian. Therefore $1\ne H\ne G$. Also, $H\lhd G$ because already the set of generators is invariant under conjugation. Therefore $G$ is not simple.
(Note that this does not even need that $G$ be finite).
A: We still need to show that $G$ has to be nilpotent. This will be proved by induction on $|G|$. Firstly, let us remark that the property "any two elements that are conjugate commute with each other" is inherited by subgroups and quotients. It is obvious for subgroups, for quotients it is more subtle: assume $N \unlhd G$, and $\bar{g}, \bar{h} \in G/N$ are conjugate. Say $\bar{h}=\bar{x}^{-1}\bar{g}\bar{x}$ for some $x \in G$. Hence, there is an $n \in N$ with $hn=x^{-1}gx$. Hence $hn$ and $g$ commute in $G$, which implies that $\bar{g}$ and $\bar{h}$ commute in $G/N$.
Now we need two lemma's that are quite standard.
Lemma 1 Let $G$ be a group, $N \unlhd G$, such that $G/N'$ and $N$ are both nilpotent. Then $G$ is nilpotent.
Proof See this site here.
Lemma 2 Let $G$ be a group, $M, N \unlhd G$, with $G=MN$ and both $M$ and $N$ abelian. Then $G$ is nilpotent.
Proof $G/N \cong M/(M \cap N)$ is abelian, so $G' \subseteq N$. By symmetry, $G' \subseteq M$. Hence, $G' \subseteq M \cap N$. But since $G=MN$, $M \cap N \subseteq Z(G)$. So $G' \subseteq Z(G)$ and $\gamma_3(G)=[G,G']=1$.
Let $G$ a counterexample of minimal order. We can assume that $G$ is not abelian. Observe that by Lemma 1, every proper normal subgroup of $G$ must be abelian. Let $M$ be a maximal normal subgroup (such an $M$ exists, $G$ is not simple, as was pointed out by Hagen) and take $x \in G - M$. Then by the property, $N:=\langle Cl_G(x) \rangle$ is a non-trivial proper abelian normal subgroup of $G$. Since $x \notin M$ and $M$ is maximal normal, we see that $G=MN$. Now apply Lemma 2 and we arrive at a contradiction.$\square$
A: As has been noted, we must have that $G$ is non-abelian, as all abelian simple groups are counterexamples.
Else, let $1\neq g\in G$ and consider $$H=\bigcap_{x\in G}x C_G(g)x^{-1} = \bigcap_{x\in G} C_G(x g x^{-1}).$$  Note that $H$ is a subgroup, and is normal by construction.  By assumption $\operatorname{class}(g)\subseteq C_G(xgx^{-1})$ for all $x$.  It follows that either $H=G$, in which case $g\in Z(G)$ and so $Z(G)$ is a proper normal subgroup, or that $H$ is a proper normal subgroup.
