$C_{S_4}(A_4)=1$

Question

I have to show that $C_{S_4}(A_4)$ is trivial.

Now we know that $$ C_{S_4}(A_4)=\{x\in S_4\;:\:yx=yx\;\;\forall y\in A_4\}=\bigcap_{y\in A_4}\{x\in S_4\;:\;yx=xy\}\;. $$

Then every element $\neq1$ in $A_4$ can be written as $(abc)$ or $(ab)(cd)$. Then $(ac)\in S_4$ doesn't commute with the last two. But I can't see in which manner this could help.

Thank you all

Answer 1

As Derek pointed out,

Let $\sigma\in S_4$ then $\sigma^{-1}(a,b,c)\sigma=(\sigma(a),\sigma(b),\sigma(c))$

The above equality is standard fact used for "under conjugation, cycle form does not change in $S_n$".

So, we want to $\sigma \in C_{S_4}(A_4)$ then, $$\sigma^{-1}(a,b,c)\sigma=(\sigma(a),\sigma(b),\sigma(c))=(a,b,c)$$ then $\sigma$ must fix $d$ and by similiar argument, it fixes $a$ , $b$ and $c$ so $\sigma$ is trivial.