# Showing the permutation of 2 elements in a symmetry group is an even permutation

Show that for every 2 elements $\alpha$ and $\beta$ in $S_{8}$, the permutation $\alpha ^{-1}\beta ^{2}\alpha$ is an even permutation.

How do I show that the above is an even permutation? I know that permutations are always either even or dd and that even permutation forms a subgroup. But I'm stuck on this one.

• Just knowing that even permutations form a subgroup isn't specific enough ($S_8$ has lots of subgroups and $\alpha^{-1}\beta^2 \alpha$ isn't going to belong to all of them). What's something specific you know about even permutations? Can you give any example of one? – Erick Wong Jun 12 '15 at 6:51
• Hint: Think of it like numbers. If $\beta$ is odd, what is the parity of $\beta^2$? Same question for even. Same questions for $\alpha$. – Joe Moeller Jun 12 '15 at 6:52

Here Notice that ${\beta}^2$ is always even permutation.

Case-1 $\alpha$ even permutation

So ${\alpha}^{-1}$ is also even permutation and hence ${\alpha}^{-1}{\beta}^2\alpha$ is even permutation

Case-2 $\alpha$ odd permutation

So ${\alpha}^{-1}$ is also odd permutation and hence both $\alpha ~\&~{\alpha}^{-1}$ are product of odd number of transpositions and ${\beta}^2$ is product of even no. of transpositions.

So ${\alpha}^{-1}{\beta}^2\alpha$ is product of odd+even+odd=even no. of transpositions. And hence ${\alpha}^{-1}{\beta}^2\alpha$ is an even permutation.

You can prove it using $alternating~map$

$\psi ({\alpha}^{-1}{\beta}^2\alpha)=\psi(\alpha^{-1})~ \psi(\beta^2) ~\psi(\alpha)=(\psi(\beta))^2=1$

which also shows that ${\alpha}^{-1}{\beta}^2\alpha$ is an even permutation