Combining Operation for $S_4$

Question

If $\sigma=(1\ 2\ 3\ 4)$, $\kappa=(1\ 2)$ for $S_4$ and I want to compute $(\sigma\kappa)^2$, does it become $\sigma^2\kappa^2 = \sigma^2$ (since $\kappa^2 = 1$), which is just $(1\ 3)(2\ 4)$?

Answer 1

In general, $(\sigma k)^2$ means $(\sigma k)(\sigma k)$, because you "multiply" the element $(\sigma k)$ two times with itself. (It's actually the group operation instead of multiplication).

In this case, $\sigma k\sigma k$ is not the same as $\sigma^2 k^2$, because the group $S_4$ is non-commutative.

Does this help?

Answer 2

Since $(\sigma\kappa)^2=\sigma\kappa\sigma\kappa$, asserting that $(\sigma\kappa)^2=\sigma^2\kappa^2(=\sigma\sigma\kappa\kappa)$ is the same thing as asserting that $\kappa\sigma=\sigma\kappa$. Well, this is not true. For instance, $\kappa\sigma$ maps $1$ into itself, whereas $\sigma\kappa$ maps $1$ into $3$.

Anyway, I think that the most natural approach for this problem is simply to compute $\sigma\kappa$ and then to square it.