# Index of certain subgroups of the free group $F_2$

Suppose that $F = \langle a,b \rangle$ is the free group on two generators and let $H=\langle X,Y\rangle$ be the subgroup of $F$ generated by $X = (ab)^k$, $k$ non-zero integer, and $Y = a$.

What is the index of $H$ in $F$ ?

We know that $H$ is a free group on two generators. When $k = \pm 1$, it is easy to see that $F = H$. But what about when $k \neq \pm 1$ ?

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Since F is generated by a and ab, which have no relations, you could replace ab by b in your definition of X. That is, F = <a,b> and H = <Y,X> = <a,b^k>. The group SL(2,Z) has two standard generators S = [0 -1|1 0] and T = [1 1|0 1], where S has order 4 and ST has order 6. While <S,T^2> has index 3 in SL(2,Z) = <S,T>, for any k > 2 the subgroup <S,T^k> has infinite index in SL(2,Z). The proof, shown to me by Vincentiu Pasol, uses the action of SL(2,Z) on primitive vectors in Z^2. Because of this result,[F:H] is infinite for k > 2. – KCd Jun 20 '11 at 20:49
The above seems worth writing as an answer instead of a comment. – David Kohler Jun 20 '11 at 21:32
As a general rule any proper subgroup of $F_2$ isomorphic to $F_2$ must have infinite index. A finite-index subgroup of a free group is always a free group of higher rank. – Jim Belk Jun 20 '11 at 23:03
Jim: a subgroup of index 1 wouldn't have higher rank and your statement is also incorrect for a free group of rank 1. – KCd Jun 21 '11 at 2:54
@KCd: Jim said "proper", so index 1 doesn't count. Also, he was speaking of $F_2$ (and he's right), not of free groups of rank 1. – A. De Luca Jun 21 '11 at 13:21

I would do this computation by thinking about covering spaces of graphs and applying Stallings's folding algorithm.

We realise $F_2$ as the fundamental group of a graph $X$ with one vertex $v$ and two edges---this is sometimes called the rose with two petals. We orient the edges, and label one by $a$ and the other by $b$. This fixes an identification $\langle a,b\rangle\equiv\pi_1(X,v)$.

Your subgroup $H$ can be thought of similarly. It is the fundamental group of a graph $Y$, which we construct as follows:

1. Fix a base vertex $*$.
2. Attach both ends of an oriented edge labelled $a$ to $*$ .
3. Attach both ends of an oriented interval, consisting of $2k$ edges labelled $a$ and $b$ alternately, to $*$.

The orientations and labels define a natural map $Y\to X$, and the image of $Y$ is your subgroup $H$.

(For more on this sort of construction see, for instance, this blog post.)

Stallings' folding algorithm is a way of turning this map into an immersion---that is, a local embedding. The algorithm is easy:

1. If two edges with the same label are both oriented into the same vertex, identify them.
2. If two edges with the same label are both oriented away from the same vertex, identify them.
3. Repeat.

At the end of this procedure, we have a new oriented, labelled graph $Y'$, and the map $Y'\to X$ is an immersion. There are essentially two possibilites:

1. Every vertex of $Y'$ has valence four. If so, then $Y'\to X$ is a covering map and $H$ is a finite-index subgroup of $F_2$. The index of $H$ is equal to the degree of the covering map, which is equal to the number of vertices of $Y'$.

2. Some vertex of $Y'$ has valence less than four. If so, then $H$ is of infinite index. (To see this, note that you can complete it to an infinite-sheeted covering space.)

In your case, you can quickly see that you need to perform exactly one fold to turn $Y$ into $Y'$. If $k=1$ then $Y'$ is isomorphic to $X$ and your subgroup $H$ is equal to $F_2$. Otherwise, $Y'$ has a vertex of valence two and $H$ is of infinite index.

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Further to Jim Belk's comment, Thereom 2.10 of Magnus, Karrass and Solitar states,

Let $F$ be the free group on $a_1, \ldots, a_n$, and let $j$ be the index of the subgroup $H$ in $F$. If both $n$ and $j$ are finite, then $H$ is a free group on $j(n-1)+1$ generators. If $n$ is infinite and $j$ is finite, then $H$ is a free group on infinitely many generators. Finally, if $j$ is infinite, then $H$ may be finitely or infinitely generated; however, if $H$ contains a normal subgroup $N$ of $F$, $N\neq 1$, then $H$ is a free group on infinitely many generators.

For the example given, plugging in the values we see that your subgroup must have index $1$ if it is of finite index.

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Just to remark that the first statement is known as Schreier's formula. – A. De Luca Jun 21 '11 at 13:42