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Let $H$ be the normal subgroup of $G$.

Is it true that $(H[G,G])/H$ is isomorphic to $[G/H,G/H]$?

If so, I want to make a surjective homomorphism $\phi\colon H[G,G]\to [G/H,G/H]$ with Kernel $H$ to prove it.

But, if I define $\phi(h[g_1,g_2])=[g_1/H,g_2/H]$ then I have some trouble.

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1  
What kind of trouble? –  mac May 14 '11 at 22:12
    
For example, does it group homomorphism? I don't think so. And, my formula doesn't depend on $h$. –  group homology May 14 '11 at 22:14
2  
but your formula is just $\phi(g)=g/H$ (restricted to $g\in H[G,G]$) –  user8268 May 14 '11 at 22:16
    
You've not defined the map for every element of $H[G,G]$, but the elements $h[g_1,g_2]$ generate this group. I guess a typical element is of the form $h[g_1,g_2][g_3,g_4]\dots[g_{k-1},g_k]$, so you can extend $\phi$ without much trouble. If $h_1,h_2\in H$ then $h_1[g_1,g_2]h_2[g_3,g_4]=h_3[g_1,g_2][g_3,g_4]$ for some $h_3\in H$ since $H$ is normal. This sort of consideration makes me think $\phi$ is probably well-defined and multiplicative. And what user8268 said looks right too, and is a much nicer way to think about it! –  mac May 14 '11 at 22:19

2 Answers 2

up vote 4 down vote accepted

The commutator subgroup is a verbal subgroup. That means that for any group homomorphism $\varphi\colon G\to K$, you have that $\varphi([G,G])\subseteq [K,K]$. If $\varphi$ is onto, then the induced map on commutator subgroups is also onto.

Applying this to the quotient map $G/H$, you have that $\pi([G,G])$ maps $[G,G]$ onto $[G/H, G/H]$. The kernel of this map is $[G,G]\cap H$, so by the isomorphism theorems we have $$\left[\frac{G}{H},\frac{G}{H}\right] \cong \frac{[G,G]}{[G,G]\cap H} \cong \frac{H[G,G]}{H}.$$ Follow through the definitions given by the isomorphism theorems, and you'll see that the isomorphism from $\frac{H[G,G]}{H}$ to $[G/H,G/H]$ is indeed given by mapping $huH$ to $\pi(u)$ (or by mapping from $H[G,G]$ onto $[G/H,G/H]$ via $hu\mapsto \pi(u)$). This is the map you give (though you only describe it for a generating set of $H[G,G]$).

The same result holds for any verbal subgroup $\mathfrak{V}(G)$: if $N$ is normal, then $$\mathfrak{V}\left(\frac{G}{N}\right) \cong \frac{\mathfrak{V}(G)}{\mathfrak{V}(G)\cap N} \cong \frac{N\mathfrak{V}(G)}{N}.$$ E.g., for the subgroup generated by the $n$th powers, $G^n$, we have $$\left(\frac{G}{N}\right)^n \cong \frac{NG^n}{N}.$$

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An easy argument certainly follows user8268's comment.

A few things should be noticed (as they might not seem totally obvious). At first since $H$ is normal in $G$, $$ [g_1H,g_2H] = [g_1,g_2]H \in G/H.$$ Now restricting the canonical epimorphism $G\to G/H$ to $H[G,G]$, we may call it $\phi$, gives us a mapping, that satisfies $$ \phi(h[g_1,g_2])= [g_1,g_2]H = [g_1H,g_2H] \in [G/H,G/H]$$ for $h\in H$ and $g_1,g_2\in G$. Since $H[G,G]$ is generated by the elements $h[g_1,g_2]$, $\phi(H[G,G])$ is certainly generated by the elements $[g_1H,g_2H]$, that generate $ [G/H,G/H]$. Hence $\phi$ is surjective and its kernel is still $H$ as it is included in $H[G,G]$.

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