Let $H\leq G$ s.t. whenever 2 elements of $G$ are conjugate then the conjugating element can be chosen from within $H$. Prove that $G'\subseteq H$.

Question

Question: Let $H\leq G$ such that whenever two elements of $G$ are conjugate, then the conjugating element can be chosen from within $H$. Prove that $G'\subseteq H$.

Solution: So, if $g_1$ and $g_2$ are conjugate then $g_1=hg_2h^{-1}$ for some $h\in H$. If we let $a\in G'$, then $a=x^{-1}y^{-1}xy$ for some $x,y\in G$. Then, $xa=y^{-1}xy$ and $ay^{-1}=x^{-1}y^{-1}x$. So, $xa$ and $x$ are conjugate, and $ay^{-1}$ and $y^{-1}$ are conjugate. So by hypothesis, $y,x\in H$ so $a\in H$.

Does this look okay? To me, it feels "clunky", but maybe it is just me.

Answer 1

Let $x,y \in G$. Then $x^{-1}$ and $y^{-1}x^{-1}y$ are conjugate so there exists $h \in H$ with $y^{-1}x^{-1}y = h^{-1}x^{-1}h$ and hence $y^{-1}x^{-1}yx = h^{-1}x^{-1}hx$.

Similarly, there exists $k \in H$ with $x^{-1}hx = k^{-1}hk$, so $y^{-1}x^{-1}yx = h^{-1}k^{-1}hk \in H$.

So $H$ is a subgroup containing all commutators, and the commutators generate $G'$, so $G' \le H$.

Answer 2

Derek's answer is definitely quick and nice. Here is another approach, proving that $H$ is normal first.

If $h\in H$ and $g\in G$ then $h$ and $h^g$ are conjugate, so there exists $k\in H$ such that $h^g=h^k\in H$. Thus $H$ is normal in $G$.

Now we prove that $G/H$ is abelian, so $H$ contains $G'$. If $x,y\in G$ then $xyx^{-1}=hyh^{-1}$ for some $h\in H$, yielding $$HxHy=Hxy=Hxyx^{-1}x=Hhyh^{-1}x=Hyh^{-1}x=HyHh^{-1}Hx=HyHx.$$