Quite a simple looking question guys,
Find the centre $Z(S_4)$ of $S_4$.
The previous part asked me to find centralizers for $S_4$. I note that $Id$ is the only element contained in everything so I am thinking the centre is just the trivial group {Id} but I am a little unsure how to formally prove.
Cheers!
On
If we denote the center of $a$ in $G$ by $C_{G}(a)$ then it holds that $$ Z(G)=\cap_{a\in G}C_{G}(a) $$
This is because the centralizer of $a$ in $G$ is the set of all the elements the commutes with $a$ and an element is in the center iff it commutes with all elements.
This gives justification to your reasoning that the center is trivial
Let $\sigma \in Z(S_4)$. Then $\sigma^{-1}(12)\sigma=(12)$. But $\sigma^{-1}(12) \sigma=(\sigma(1) \sigma(2))$. Hence either $\sigma$ fixes $1$ and $2$, or $\sigma$ flips $1$ and $2$. Suppose the latter, then $\sigma^{-1}(123)\sigma=(1\sigma(3)2)\neq (123)$, contradicting the fact that $\sigma$ is central. So $\sigma$ fixes $1$ and $2$. Now similary, working with $(34)$ and $(134)$, one shows that $\sigma$ fixes $3$ and $4$. So after all, $\sigma$ is the identity of $S_4$.