## Characterize the conjugacy classes of Alt(n)

Let $K \subseteq S_n$ be a conjugacy class and suppose that $K \subseteq A_n$.

1. Show that $\sigma \in S_n$ does not commute with any odd permutation if and only if the cycle type of $\sigma$ consists of distinct odd integers.
2. Deduce that $K$ is a union of two $A_n$-conjugacy classes if and only if the cycle type of $K$ consists of distinct odd integers.

[Hint: Assume first that $\sigma \in K$ does not commute with any odd permutation. Observe that $\sigma$ commutes with each cycle in its own cycle decomposition, so that each cycle must have odd length. If two cycles have the same odd length $k$, find a product of $k$ transpositions which interchanges them and commutes with $\sigma$. Conversely, if the cycle type of $\sigma$ consists of distinct integers, prove that $\sigma$ commutes only with the subgroup generated by the cycles in its cycle decomposition.]

We begin with a remark.

Remark: If $\sigma \in S_n$ is a cycle, then $C_{S_n}(\sigma) = \langle \sigma, S_{F(\sigma)}$, where $F(\sigma)$ is the set of fixed points of $\sigma$ and $S_{F(\sigma)}$ is naturally identified with a subgroup of $S_n$.

Now to the main result.

$(\Rightarrow)$ Suppose $\sigma \in S_n$ does not commute with any odd permutation. Write $\sigma = \alpha \tau$, where $\alpha$ is some cycle in the decomposition of $\sigma$ and $\alpha$ commutes with $\tau$. (Say $\tau$ is the rest of the cycle decomposition of $\sigma$.) Note that $\alpha \sigma = \alpha \alpha \tau = \alpha \tau \alpha = \sigma \alpha$, so that $\alpha$ commutes with $\sigma$. By our hypothesis, then, $\alpha$ is not an odd permutation. Hence $\alpha$ has odd length. Since $\alpha$ was an arbitrary cycle in the decomposition of $\sigma$, the cycle type of $\sigma$ consists of odd integers.

Now suppose some cycles $\alpha$ and $\beta$ in the decomposition of $\sigma$ have the same length $k$, and write $\sigma = \alpha \beta \theta$ for some $\theta$ which commutes with both $\alpha$ and $\beta$. (Again, say, the rest of the cycle decomposition of $\sigma$.) Now $\alpha$ and $\beta$ are conjugate in $S_n$, and moreover are disjoint. Write $\alpha = (a_1\ a_2\ \ldots\ a_k)$ and $\beta = (b_1\ b_2\ \ldots\ b_k)$; that is, define $a$ and $b$ such that $\alpha(a_i) = a_{i+1}$ and $\beta(b_i) = b_{i+1}$ for all $1 \leq i \leq k$, where indices are taken mod $k$. Let $\tau_i = (a_i\ b_{i+1})$ for each $1 \leq i \leq k$, and let $\tau = \tau_1 \tau_2 \cdots \tau_k$; note that $\tau$ commutes with $\theta$. Moreover, by a previous exercise, we have $M(\tau \alpha \tau^{-1}) = \tau[M(\alpha)] = M(\beta)$ by our definition of $\tau$. Now for each $i$, we have $(\tau \alpha \tau^{-1})(b_i) = (\tau \alpha)(a_{i-1}) = \tau(a_i) = b_{i+1} = \beta(b_i)$, with indices taken mod $k$. Hence $\tau \alpha \tau^{-1} = \beta$. Moreover, since $\tau$ is a product of disjoint 2-cycles, $\tau^{-1} = \tau$. Finally, note that $\tau$ is an odd permutation, being an odd product of commuting transpositions. Now $\tau \sigma = \tau \alpha \beta \theta = \beta \tau \beta \theta = \beta \alpha \tau \theta = \alpha \beta \theta \tau = \sigma \tau$, so that $\tau$ commutes with $\sigma$, a contradiction. So no two cycles in the decomposition of $\sigma$ have the same length.

Thus the cycle type of $\sigma$ consists of distinct odd integers.

$(\Leftarrow)$ Suppose that the cycle type of $\sigma$ consists of distinct odd integers. Then $\sigma$ fixes at most one element; that is, $|F(\sigma)| \leq 1$. By the remark above, $C_{S_n}(\sigma) = \langle \sigma, S_{F(\sigma)} \rangle$. Now $S_{F(\sigma)}$ consists of only the identity; hence $C_{S_n}(\sigma) = \langle \sigma \rangle$. Because $\sigma$ is a product of even permutations, $C_{S_n}(\sigma)$ consists only of even permutations. In particular, $\sigma$ does not commute with any odd permutation.

Now for the second point.

Let $K \subseteq A_n$ be a conjugacy class and let $\sigma \in K$.

Note that $K$ consists of two $A_n$-conjugacy classes if and only if $\sigma$ does not commute with any odd permutation by §4.3 #20, if and only if the cycle type of $\sigma$ consists of distinct odd integers by the above argument.

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### Comments

• Gobi Ree  On December 12, 2011 at 12:33 am

In ($\Rightarrow$), how about defining $\tau=(a_1\ b_1) \cdots (a_k\ b_k)$? Then the conjugation of $\tau$ changes $\alpha$ with $\beta$ so that $\tau \sigma \tau^{-1}=\sigma$.

• nbloomf  On December 15, 2011 at 11:08 am

I like it.