Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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!

share|improve this answer
    
Ah, too bad. Thanks for the example! –  Kevin Dec 12 '11 at 2:42
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.