I am working on the following exercises but am having some difficulty starting.
let $N \unlhd S_n$ with $n \geq 5$.
(a) Let $G$ be a group and let $H \unlhd G$ be a normal subgroup isomorphic to $C_2$. Show that $H < Z(G)$.
(b) Suppose that $N \cap A_n \neq \{id\}.$ Show that $A_n \subset N$ and conclude that $N = A_n$ or $N = S_n$.
(c) Suppose that $N \cap A_n = \{id\}.$ Show that $N$ is isomorphic to a subgroup of $C_2$.
(d) Show that if $n \geq 3$ then $Z(S_n) = \{id\}$, and conclude that in case (c) we must have $N = id$.
My strategy for (a) is to note that $C_2$ is abelian and so $h$ must be abelian as well. Therefore $H < Z(G)$.
For part (a): You have $H=\{1,h\}$ and $gHg^{-1}=H$ for all $g\in G$. What are the possibilities for $ghg^{-1}\in gHg^{-1}=H$?
For part (b): If $N\lhd G$ and $H\leq G$, then $N\cap H\lhd H$. Applying this to $H=A_n$, what can you say?
For part (c): For $N\cap A_n=\{1\}$, $NA_n$ properly contains $A_n$, so $NA_n=S_n$. Therefore, $|NA_n|=n!$. Can you compute $|NA_n|$ in terms of $|N|$? Then use part (a).
For part (d): Let $\sigma\in Z(S_n)\backslash\{1\}$. Then, there exist $i\neq j$ such that $\sigma(i)=j$. Since $n\geq 3$, there exists $k\in\{1,\ldots,n\}\backslash\{i,j\}$. Let $\tau$ be any permutation such that $\tau(i)=i$ and $\tau(j)=k$. Why do we have $\tau\sigma\neq \sigma\tau$?