I'd appreciate comments on the validity of these attempted proofs. Thanks.
Let $a$ be an $n$-cycle in $S_n$.
a) Show that the centralizer of $a$ in $S_n$ is $\langle a \rangle$.
b) Assume that $n$ is odd. Show that $A_n$ contains an $n$-cycle $b$ that is not conjugate to $a$ in $A_n$.
a) Since the number of $k$-cycles in $S_n$ is given by $\frac{n!}{(n-k)!k}$, the number of $n$-cycles in $S_n$ is $(n-1)!$. Since two elements of $S_n$ are conjugate iff they have the same cycle type, $|\mathrm{cl}(a)|=(n-1)!$ (where $\mathrm{cl}(a)$ is the conjugacy class of $a$). But $|\mathrm{cl}(a)|=|S_n:C_{S_n}(a)|$, or $|C_{S_n}(a)|=\frac{|S_n|}{|\mathrm{cl}(a)|}=\frac{n!}{(n-1)!}=n$. Since $\langle a \rangle \le C_{S_n}(a)$, it follows that $\langle a \rangle = C_{S_n}(a)$.
b) since $n$ is odd, all $n$-cycles are contained in $A_n$, so $A_n$ contains $(n-1)!\,\,$ $n$-cycles. Since, for $a \in S_n$, $C_{A_n}(a)=C_{S_n}(a) \cap A_n = \langle a \rangle \cap A_n$, we have $|\mathrm{cl}_{A_n}(a)|=|A_n:C_{A_n}(a)|=\frac{\frac{1}{2}n!}{n}=\frac{n!}{2n}=\frac{1}{2}(n-1)!$, and so $a$ cannot be conjugate to all $(n-1)!$ $n$-cycles in $A_n$ and there exists such an element $b$.