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Suppose $G$ is a perfect group, i.e. it is equal to its commutator subgroup, or equivalently has trivial abelianization. If $H<G$ is a finite index subgroup, can anything be said about the abelianization of $H$? For instance, must it be finite?

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up vote 4 down vote accepted

The hyperbolic triangle group $G=\langle a,b\ |\ a^3=b^4=(ab)^5=1\rangle$ is perfect (check its abelianization). $G$ maps onto the finite simple group $A_6$, by sending $a$ to $(125)$ and $b$ to $(1436)(25)$. The kernel $K$ of this map has index $360$, and has abelianization $\mathbb{Z}^{80}$. You can do all this by hand if you like using Fox Calculus, but GAP can do it much quicker!

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Ah, too bad. Thanks for the example! – Kevin Dec 12 '11 at 2:42

Here is a different kind of answer:

If H is a finite perfect group and V is a non-trivial irreducible H-module, then the semi-direct product $G=H\ltimes V$ is a perfect group with an abelian subgroup V of finite index.

Indeed, $[G,G]=[H,H][H,V][V,V]=H[H,V]$, but since V is irreducible and $[H,V]$ is a nonzero submodule of V, $[H,V]=V$ and $[G,G]=HV=G$, so that G is perfect.

In particular, the affine special linear group, $\operatorname{ASL}(n,K)$, is such a group for any infinite field K.

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