# 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.

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It sounds to me from your question that you are not clear about what $S_n$ is. Do you need help in that regard? – Zev Chonoles Feb 10 '13 at 1:01
$S_n$ is the group of permutations {1,2,3,...n}. So how can we compute from here? – Faye Feb 10 '13 at 1:02
What have you tried? For example, I bet you can do, say, $S_3$ and $S_4$ by hand... does that help you see what the general case might look like? – Dylan Wilson Feb 10 '13 at 1:06

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 $$Z(G) = \{ x \in G : y^{-1} x y = x \text{ for all y \in G} \}.$$

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thanks Andreas for the great answer you have provided. – Faye Feb 21 '13 at 20:29
@Faye, you're welcome. – Andreas Caranti Feb 21 '13 at 20:33

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 $S_n$: $|S_n|=n!$

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Thanks for your help. – Faye Feb 10 '13 at 1:11
@CrystalSeluini You're welcome, I updated my answer to explain what $Z(G)$ is, as you asked it in a comment. Good luck! – MyUserIsThis Feb 10 '13 at 1:17
Thanks, it is very much appreciated! Cheers. – Faye Feb 10 '13 at 8:14
@CrystalSeluini I'm simply curious if based on this general answer have you succeded to find the answer to your own question. (Btw, meanwhile you have got two more answers that deserve more IMO to be upvoted and/or accepted than this one.) – user26857 Feb 10 '13 at 16:50

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 $n>3$.

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thanks you for your help Ludolila. – Faye Feb 21 '13 at 20:31