Suppose $G$ is a perfect group, i.e. it is equal to its commutator subgroup, or equivalently has trivial abelianization. If $H
Abelianization of finite index subgroup of a perfect group
2 Answers
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!
-
0Ah, too bad. Thanks for the example! – 2011-12-12
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.