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.

How can I use coverings of graphs to show that if $G$ is a finitely generated free group and $H$ is a subgroup of finite index, then $H$ is finitely generated.

I've seen this done without using coverings of graphs, but I am curious to see how it's done otherwise.

Can anyone help?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let $G$ be generated by $n$ elements; then $G$ is the fundamental group of the wedge of $n$ circles (call this space $X$). A finite-sheeted covering of $X$ corresponds to a finite-index subgroup $H$. But if the covering $Y$ associated to $H$ is $d$-sheeted, then $Y$ has $d$ vertices and $nd$ edges. $Y$ is thus a finite graph, which therefore only has finitely many loops, and so $H$ is finitely generated (indeed, by at most $nd$ elements).

Further work shows contracting a maximal tree in $Y$ doesn't change $\pi_1(Y)\cong H$, so that $H$ is in fact free, on $nd-(d-1)$ elements.

share|improve this answer
    
That worked out much nicer than I thought it would. Thank you for the amazing answer! –  josh Jun 3 '12 at 16:40

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.