Prove that there doesn't exist any normal subgroup $H$ such that $S_5/H $ is isomorphic to $S_4$ My attempt :
Order of the group $S_5$ is $5!$ so by Lagrange's theorem the order of the group $H$ should be $5$.So it must be a cyclic group generated by a $5$ cycle then let $(a_1 a_2 a_3 a_4 a_5)$ be a generator of the subgroup. Then for any element $g \in S_5$ ,$g(a_1..a_5)g^{-1} \in H$ , then $g(a_1)..g(a_5) \in H$.Now, all the elements of $H$ are $5$ cycles. Now if we choose $g$ in such a manner that $g(a_1)..g(a_5)$ becomes a two cycle then I am done. So I choose $g(a_1)=a_2$,$g(a_2)=(a_1)$..Is this OK?  I don't think it is right where am I going wrong?

You are on the right track, you've shown that such a normal subgroup would need to be generated by a $5$ cycle in $S_5$, so now we need to show that the subgroup generated by a $5$ cycle in $S_5$ can't be normal. Conjugating a $5$ cycle will always yield a $5$ cycle (since conjugation preserves the order of elements of a group), so we'll need to work a little harder. There are a few ways to see this, but I'll give a hands on proof. Its enough to show that from any $5$ cycle, we can find something in the subgroup generated by its conjugates that isn't a $5$ cycle, since then we can't have any group generated by a $5$ cycle being normal. So if we have $(abcde)$ is our $5$ cycle, conjugating by the transposition $(ab)$ yields the $5$ cycle $(bacde)$, and $(abcde)(bacde)^{-1}=(cba)$, which has order $3$, so gives something bigger than our hypothetical normal subgroup of order $5$. Thus, no normal subgroup of order $5$ can exist.
