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 have almost finished the proof. So $|H|=5$, $H$ consists of 4 5-cycles and the identity element. Since there are $24$ cycles of length  $5$ in $S_5$ there exists a $5$-cycle $c$ not in $H$. Since any two $5$-cycles are conjugate $S_5$, there is a conjugation taking a cycle from $H$ to $c$. This is a contradiction since $H$ is normal.
