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Here is a paraphrased version of problem 4B.4 in Isaacs's Finite Group Theory:

Let $G$ be a group and $X,Y$ subgroups of $G$, such that $Y$ centralizes $[X,Y]$. If $X$ is normal in $G$, show $[X,Y]$ is abelian.

Now of course we can assume $G=\langle X,Y\rangle$, and thus $K=[X,Y]$ is normal in $G$. But then $C_G(K)$ is normal in $G$ too, and in the quotient $G/C_G(K)$, $\bar{K}=[\bar{X},\bar{Y}]=[\bar{X},\bar{1}]=\bar{1}$, so $K\subset C_G(K)$ and $K$ is abelian.

So my question is:

Why do we need to assume $X$ is normal in $G$?

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    Yes, like everone else, I agree. It isn't even really necessary to reduce to $[H,K] \lhd G$. Because $K$ normalizes $[H,K]$, we see that $y^{x}$ centralizes $[H,K]$ for each $y \in H$ and $x \in K$. Hence $y^{-1}y^{x}$ centralizes $[H,K]$ for all such $x,y$. But $[H,K]$ is generated by such elements.2011-07-15

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