G = $S_{n}$ Suppose H ≤ Sn and [$S_{n}$ : H] = 2
1) Let $\pi: G\rightarrow G/H$ be the quotient map. Show that if ker(π) contains one 2-cycle, then it must contain every 2-cycle.
2) (1) implies either ker(π) contains every 2-cycle or it contains no 2-cycle. Argue that the former is impossible.
This is what I've done so far:
For 1): Since H$\triangleleft$G (by the definition of the map π that sends g to gH and is a surjective homomorphism with kernel H), if it contains one 2-cycle, then it contains all of them: let $\sigma \in S_{n}$, then $\exists \alpha \in H$ ( $\sigma$ is a 2-cycle) : $\alpha ' = \sigma\alpha\sigma^{-1}$ Since H is normal, $\alpha'$ is also in H. Therefore, H contains every 2-cycle
I don't really understand how to approach 2). And why does it imply that ker(π) contains every 2-cycle or it contains no 2-cycle. I don't see how it follows from 1).
The $2$-cycles together generate $S_n$, so the only subgroup containing all $2$-cycles is $S_n$ itself. Since the subgroup is of index $2$, it therefore can't contain all $2$-cycles.
As to the rest of your question, the subgroup either contains at least one $2$-cycle or no $2$-cycles, and if it contains at least one then it contains them all.