Subgroup of $\mathbb{Z}\ast \mathbb{Z}$ I want to see $\mathbb{Z}\ast\mathbb{Z}$ has a subgroup of index 4 which is not normal and I want to calculate the rank of this subgroup.
One way is to classify all the subgroups of $\mathbb{Z}\ast\mathbb{Z}$. However, I want to know if there is easier way to see this.
 A: Somewhat interestingly, many results about free products are proved geometrically, and if I'm being honest, this technique is so useful I'm not familiar with purely algebraic ways to do what you're asking.
As such, this answer is going to rely on some Algebraic Topology, which you may have to take on faith depending on your background.
The idea is based on the following observations/theorems:

*

*$\mathbb{Z} \ast \mathbb{Z}$ is the
fundamental group
of a wedge of $2$ circles.

*Subgroups of a fundamental group are in bijective correspondence with
covering spaces
of the underlying space.

*The subgroups of index $n$ are in bijection with the $n$-sheeted covering spaces.

*The normal subgroups are in bijection with the "maximally symmetric" covering spaces.

So, to find a non-normal subgroup of index $4$, we want to find a covering of the wedge of $2$ circles with $4$ sheets that isn't symmetric.
The wedge of $2$ circles looks like this:

(the loops $a$ and $b$ correspond to the generators of $\mathbb{Z} \ast \mathbb{Z}$)
We want a covering space (for our purposes, another graph) which has $4$ loops corresponding to $a$ and $4$ loops corresponding to $b$ (that gives $4$-sheetedness). We also want to know that some (labelled) edges cannot be taken to some other edges (of the same label) by a graph automorphism. This is what is meant by "not maximally symmetric" (formally, we want to know the group of (labelled) graph automorphisms isn't transitive).
I would draw one, but Hatcher actually provides an example on page 58 of his Algebraic Topology book (which is freely available online):

Figuring out exactly how we get this subgroup from this graph, or why this graph works, are the fundamentals of an algebraic topology class, so I won't try to describe what's going on in any detail. The quick version is

*

*There are $4$ edges labelled $a$ and $4$ edges labelled $b$. So this is a $4$-sheeted covering, and its fundamental group has index $4$ in $\mathbb{Z} \ast \mathbb{Z}$.


*If we start at the thick point and we follow $ab$, we get back where we started. However, if we look at the bottom right point instead, then $ab$ takes us to the top left point (so we don't get back where we started). This witnesses the "not maximally symmetric"ness that we need to show that the subgroup we get isn't normal.
So the subgroup that we get (which, loosely, corresponds to looking at all the ways to get from the thick point to itself):
$$\langle a^2, b^4, ab, ba^2b^{-1}, bab^{-2} \rangle$$
is non-normal of index $4$. As desired.

Edit: I totally forgot you also want the rank.
You might suspect this subgroup is of rank $5$, seeing as that's how many generators are listed. That suspicion is correct, as shown by another geometric fact:
The rank of a group that you get in this way is $1 - \chi$ where $\chi$ is the euler characteristic of the graph. Here $\chi = -4$ is the number of vertices minus the number of edges, so the rank is
$$1 - \chi = 1 - (-4) = 5.$$

I hope this helps ^_^
