Thread: Finding subgroups of index 2 of the free group.

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.

Originally Posted by 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

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.

Eh?

here, we have a happy accident: any subgroup of index 2 is automatically normal. so it's the kernel of a surjection .

well, that makes it easy: what are our choices for possible images of the generators a and b?

EDIT: your first instincts weren't that bad: look at the pictures on page 3 here:

http://www.math.oregonstate.edu/~mat...ngs2006/NS.pdf