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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?

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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.

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

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