For each $n \ge 1$ compute $Z(S_n)$ Can someone please help me on how to compute $Z(S_n)$ for each $n \ge 1$?
Does this basically mean compute $Z(1), Z(2), \ldots$? Please hint me on how to compute this.
Thanks in advance.
 A: Hint: $S_n$ denotes the symmetric group over a set of $n$ elements. It's the group of all posible permutations, so you have to find $Z(S_1),Z(S_2),...$ so you have to find the permutations that commute with every other permutations. That is the definition of $Z(G)$:
$$Z(G)=\lbrace g\in G,ga=ag\;\;\forall a\in G\rbrace$$
So it's all the elements of the group that commute with ALL members of the group. It's a generalization of the centralizer of a subgroup: if you have $H\subset G$, being $G$ a group and $H$ a subgroup, then the centralizer of $H$ in $G$ is:
$$C_G(H)=\lbrace g\in G, gh=hg\;\;\forall h\in H\rbrace$$
So the center of a group, $Z(G)$ is the centralizer of $G$ on $G$: $C_G(G)$
There are some trivial cases for little values of $n$, for example for $n=2$ the group is abelian so $Z(S_2)=S_2$. Remember the order of $S_n$: $|S_n|=n!$
A: Hint: I think you should start with $S_3$. Just fix an arbitrary permutation and see what it means for it to commute with all the other permutations in the group. The same idea works for all $n>3$.
A: An elementary, but very useful fact in this context is the following. 
You should know that every element of $S_{n}$ can be written as the product of disjoint cycles in an essentially unique way. Now if you take the conjugate of the cycle $(i_1, i_2, \dots, i_k)$ by $\sigma \in S_{n}$, you obtain
$$
(i_1, i_2, \dots, i_k)^\sigma = (i_1\sigma, i_2\sigma, \dots, i_n\sigma),
$$
where I am denoting by $i \sigma$ the value of $\sigma$ on $i \in \{ 1, 2, \dots , n \}$.
This can be used here, as for a group $G$ you clearly have
$$
Z(G) = \{ x \in G : y^{-1} x y = x \text{ for all $y \in G$} \}.
$$
