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?



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.


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