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This was a qualifying exam question. I'm not sure I even understood it, so I want to see if my reasoning is valid. The question is:

Let H be a subgroup of G. Suppose that for any two elements a,b that are conjugate in G, we can choose an element h $\in H$ so that a = $hbh^{-1}$. Show that the commutator G' is contained in H.

I proceed as follows: First I show H is normal, then I show G/H is abelian. Let h $\in H$. Then $ghg^{-1}$ is conjugate to h for all g, so $ghg^{-1} = khk^{-1}$ for some k $\in H$. So $ghg^{-1}$ is in H, so H is normal in G.

Next, let a $\in G$. Then again, for all g $\in G$, $gag^{-1} = hah^{-1}$ for some h $\in H$. Also, $a^{-1}hah^{-1} = khk^{-1}h^{-1}$ for some k $\in H$. So $aH = hah^{-1}H = gag^{-1}H$. So G/H is commutative, and so G' $\leq$ H.

Is there an error in this logic?

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To answer the immediate question, I see no error in your logic. –  Geoff Robinson Aug 23 '11 at 17:43
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up vote 4 down vote accepted

You never used $k$; but otherwise, the argument seeems fine to me. However, you don't actually need to do anything other than what you did at the end:

$[G,G]$ is generated by all elements of the form $[x,y]=x^{-1}y^{-1}xy$ with $x,y\in G$.

Pick $x,y\in G$. Then $[x,y] = (x^{-1}y^{-1}x)y = (h^{-1}y^{-1}h)y=[h,y]$ for some $h\in H$. And $[h,y] = h^{-1}y^{-1}hy = h^{-1}(y^{-1}hy) = h^{-1}k^{-1}hk = [h,k]$ for some $k\in H$. Thus, $[x,y]=[h,k]\in [H,H]\subseteq H$. Since all generators of $[G,G]$ are in $H$, $H$ contains the commutator subgroup of $G$.

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