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**ILikeDaisies** I want to show the free group of two generators has exactly three subgroups of index 2. I was told to consider homomorphisms from $\displaystyle F_{2} \rightarrow \mathbb{Z} $

I know $\displaystyle F_{2}$ is the same as going round a wedge union of two circles - I take the paths around the circles to be a and b. I was considering a map a to 0, b to 1, then we have a homomorphism from $\displaystyle F_{2} \rightarrow \mathbb{Z} $ (right?), and from the first isomorphism theorem, the kernel of this homomorphism should give a normal subgroup of $\displaystyle F_{2}$ - so we get a subgroup with generators $\displaystyle a, b^{2}$.

But the idea of this question is to find double coverings of the wedge union, which $\displaystyle ab^{2}$ clearly isn't.

Even if I could find subgroups index 2, I don't see how I would know when I've found all of them.

My group theory is so bad, I would appreciate any help on this.