# $C_{S_4}(A_4)=1$

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

• Every element that centralizes $(a,b,c)$ must fix $d$. Similarly, every element that centralizes $(a,b,d)$ must fix $c$, etc. So anything that centralizes all of $A_4$ must fix all four points and hence is the identity. Commented Jun 4, 2014 at 17:59
• Thanks Derek. I think I can prove it by showing that, given $(abc)$, if $x\in S_4$ moves $d$, then it would sends it wlog to $a$. Hence $(abc)x$ would move $d$ to $b$ and $x(abc)$ would move $d$ to $a$, hence $x$ and $(abc)$ couldn't commute. Am I right?
– Joe
Commented Jun 4, 2014 at 18:16
• @Joe That sounds right to me. Commented Nov 28, 2017 at 17:36

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.