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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

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    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Commented Jun 4, 2014 at 17:59
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    $\begingroup$ 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? $\endgroup$
    – Joe
    Commented Jun 4, 2014 at 18:16
  • $\begingroup$ @Joe That sounds right to me. $\endgroup$ Commented Nov 28, 2017 at 17:36

1 Answer 1

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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.

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