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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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    [Moderator's note, @user12484 's comment was converted from a misused "answer", which unfortunately resulted in the last sentence being cut off. The following is the full last sentence as written by the user.] Why is $\langle a,(ab)^k\rangle$ is a proper subgroup of $\langle a,b\rangle$ when $k\neq \pm 1$?2019-03-08

2 Answers 2

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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_.2011-06-21