0
$\begingroup$

How would I find the generators for a non-normal index 3-subgroup of the free group $\langle a,b| - \rangle$ ?. I know that a finitely generated free group can be realised as the fundamental group of a wedge of circles, so it seems I should be looking at the covering space of this bouquet. If I consider $G = S_{3}$ and $H = \{(1),(12)\}$ I know that is not-normal of index 3, but then what?

Thanks

$\endgroup$
1
$\begingroup$

Some ideas:

Define a set map $${a,b}\to S_3\;,\;\;a\mapsto (12)\;,\;\;b\mapsto (123)\;$$ and extend it to a group homomorphism $\;\pi:F(a,b)\to S_3\;$ using the universal property of free groups. If the homomorphism's kernel is $\;N\;$ , then you have

$$F(a,b)/N\cong S_3.$$

Well, by the correspondence theorem, what you want is $\;K:=\pi^{-1}(H)\le F(a,b)\;$ ...

This doesn't seem to help much about the generators of $\;K\;$, though you know there is a finite number of them, but now translating all the above to topology can help with that.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.