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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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There is no need to assume that $X$ is normal in $G$ and your reasoning is solid. There are certainly a few "errors" in Isaacs (if I remember correctly from when I read it). I would not be suprised if this is one of them. (Of course, strictly speaking, this is not an error ...) I have actually solved most of the problems in chapters 1-4 of Isaacs and have those solutions in a PDF file somewhere; I will look to see what my solution was for this problem. –  Amitesh Datta Jul 15 '11 at 5:58
    
@Steve: Just for my understanding, why is $C_G(K)$ normal in $G$? (Sorry if it's obvious, but I'm trying to learn.) –  Jesse Madnick Jul 15 '11 at 6:09
    
@Amitesh Datta: yes, I have come across several errors, but usually they are simple misprints or leaving a key word out, etc. Just to be sure, I consider this to be an excellent book! –  user641 Jul 15 '11 at 6:48
    
@Jesse: Dear Jesse, do you accept that $K$ is normal in $G$? If so, then we have that $\textbf{C}_G(K)$ is normal in $\textbf{N}_G(K)=G$ (since $K$ is normal in $G$, $\textbf{N}_G(K)=G$ by definition of the normalizer). In fact, this is a general result: if $H$ is a subgroup of $G$, then $\textbf{C}_G(H)$ is normal in $\textbf{N}_G(H)$. I will leave the proof as an exercise. (Hint: define a map $f:\textbf{N}_G(H)\to \text{Aut}(H)$ by the rule $g\mapsto \sigma_g$ where $\sigma_g(h)=g^{-1}hg$ for all $h\in H$. Prove that $f$ is a homomorphism and that the kernel of $f$ is $\textbf{C}_G(H)$.) –  Amitesh Datta Jul 15 '11 at 6:48
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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. –  Geoff Robinson Jul 15 '11 at 21:09

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