Disjoint cycle decomposition in $A_n$

Question

Please help me to prove the following result:

Show that if the disjoint cycle decomposition of $\sigma\in A_n$ includes a cycle of even lenght or two cycles of the same odd lenght then $C_{S_n}(\sigma)\not \subset A_n$

Thanks

Answer 1

An outline, with some parts for you to complete . . .

Supose first that in the disjoint cycle representation of $\sigma$, there is a cycle $c$ of even length. Since $c$ is disjoint from the cycles of $\sigma$ other than $c$, and since $c$ commutes with itself, it follows that $c$ commutes with $\sigma$. Since $c$ is a cycle of even length, it follows that $c$ is an odd permutation, hence $c \notin A_n$.

Next, suppose that in the disjoint cycle representation of $\sigma$, there are two cycles $a,b$ of the same length, $m$ say.

Show that $ab=c^2$, where $c$ is a cycle of length $2m$ whose entries are comprised of the entries of $a$ and $b$.

Since $c$ is disjoint from the cycles of $\sigma$ other than $a,b$, and since $c$ commutes with $ab$, it follows that $c$ commutes with $\sigma$. Since $c$ is a cycle of even length, it follows that $c$ is an odd permutation, hence $c \notin A_n$.

To help with your questions from the comments about how to find $c$ . . .

Suppose $a,b$ are disjoint $m$-cycles given by $$a = (x_1\;x_2\;...\;x_m)$$ $$b= (y_1\;y_2\;...\;y_m)$$ Then verify that the $2m$-cycle $$c = (x_1\;\,y_1\;\,x_2\;\,y_2\;\,...\;\,x_m\;\,y_m)$$ is such that $c^2=ab$.