Eh?I want to show the free group of two generators has exactly three subgroups of index 2. I was told to consider homomorphisms from
I know 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 (right?), and from the first isomorphism theorem, the kernel of this homomorphism should give a normal subgroup of - so we get a subgroup with generators .
But the idea of this question is to find double coverings of the wedge union, which 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.