1
$\begingroup$

Let $\sigma\in A_n$. Show that all elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$ if and only if $\sigma$ commutes with an odd permutation.

Hint: Use the previous proven fact: Assume $H$ is normal subgroup of $G$, $\mathcal{K}$ is a conjugacy class of $G$ contained in $H$ and $x\in\mathcal{K}$. Prove that $\mathcal{K}$ is a union of $k$ conjugacy classes of equal size in $H$, where $k=|G:HC_G(x)|$. Deduce that a conjugacy class in $S_n$ which consists of even permutations is either a single conjugacy class under the action of $A_n$ or is a union of two classes of the same size in $A_n$.

Proof ($\Rightarrow$): Let $\sigma\in A_n$. Assume that all elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$. We need to show that $\sigma$ commutes with an odd permutation. By our assumption we have for all $x\in \sigma S_n\sigma^{-1}$, $\sigma\in \sigma A_n \sigma^{-1}$

Proof ($\Leftarrow$): Let $\sigma\in A_n$. Assume that $\sigma$ commutes with an odd permutation. We need to show all the elements in the conjugacy class of $\sigma$ in $S_n$ (i.e. all elements of the same cycle type as $\sigma$) are conjugate in $A_n$.

Let $\tau$ be an odd permutation in $S_n$. We know that $\sigma$ commutes with $\tau$. i.e. $$\sigma\tau=\tau\sigma \iff \tau\sigma\tau^{-1}=\sigma.$$ Let $\rho$ be an element of the conjugacy class of $\sigma$ in $S_n$

I have tried to do something for each direction but I don't see how the hint is going to help me. Can I get any tips on finishing each direction?

$\endgroup$
  • $\begingroup$ Another solution using the given exercise. ⇐: Suppose that $\sigma$ commutes with an odd permutation. We show that its conjugacy class in $S_n$ cannot split in $A_n$ and we are done, since there are only two options. If that happens, by the orbit-stabilizer theorem we get: $|S_n:C_{S_n}(\sigma)| = 2|A_n:C_{A_n}(\sigma)|$ and then by opening the definition of these indexes, we get $|C_{S_n}(\sigma)|=|C_{A_n}(\sigma)|$ and therefore $C_{A_n}(\sigma)=C_{S_n}(\sigma)$, but this condradicts the fact that $\sigma$ commutes with and odd element. $\endgroup$ – math.h Apr 1 '18 at 1:27
2
$\begingroup$

It's not very hard to prove without using the hint:

  • $ \Rightarrow \,:$ Suppose a permutation in the conjugacy class in $S_n$ of $\sigma$ is also in the conjugacy class of $\sigma$ in $A_n$, i.e. a permutation $ \tau \sigma \tau^{-1} $, $\tau \in S_n\smallsetminus A_n$ can be written as $\alpha \sigma \alpha^{-1} $, $\alpha \in A_n$. Then $$\tau \sigma \tau^{-1} = \alpha \sigma \alpha^{-1} \iff (\alpha^{-1}\tau )\sigma= \sigma (\alpha^{-1}\tau ),$$ multiplying both sides by $\alpha^{-1}$ on the left and by $\tau$ on the right. This means $\sigma$ commutes with $\beta=\alpha^{-1}\tau $, which is an odd permutation.
  • $ \Leftarrow \::$ If $\sigma$ commutes with an odd $\beta$ and $\tau \sigma \tau^{-1} $ is the conjugate of $\sigma$ by an odd permutation, then \begin{alignat}{2} \tau \sigma \tau^{-1}&= \tau (\beta \sigma \beta^{-1})\tau^{-1}&&\qquad\text{since $\beta$ and $\sigma$ commute}\\ &=(\tau \beta) \sigma (\beta^{-1}\tau^{-1})&&=(\tau \beta) \sigma (\tau \beta)^{-1} \end{alignat} As $\beta$ is odd, we've finished since $\tau \beta$ is even
$\endgroup$
0
$\begingroup$

If no odd permutation commutes with $\sigma$, then $H=C_{S_n}(\sigma)\leq A_n$, and so half of the cosets of $H$ lie in $A_n$ and the other half in $S_n\backslash A_n$. So the orbit of $A_n$ containing $\sigma$ contains just half of the conjugacy class $\mathcal{K}$, and $\mathcal{K}$ splits into two classes of equal size.

On the other hand, suppose that $H$ contains an odd permutation. Then $C_{A_n}(\sigma)=C_{S_n}(\sigma)\cap A_n$ is a subgroup of index 2 in $H$. Now the size of the $A_n$-conjugacy class is the index of the stabilizer, which is $$|A_n:C_{A_n}(\sigma)|=\dfrac{|S_n:C_{A_n}(\sigma)|}{2}=|S_n:C_{S_n}(\sigma)|=|\mathcal{K}|.$$ So $\mathcal{K}$ is a conjugacy class in $A_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.