Simplify the following expression in $\Bbb Z_4\times S_4$: $([2],(123))^{-1}([1],(24))([2],(123)).$

Question

Simplify the following expression in $\Bbb Z_4\times S_4$: $$([2],(123))^{-1}([1],(24))([2],(123)).$$

I believe that I start out finding the inverse of $([2],(123))^{-1}$, which I believe is $([2],(132))$

What would I do next?

Answer 1

In groups of the form $G \times H$,we have $$(a,b) \star (c,d)=(a \star_{G} c, \,\, b\star_H d),$$ where $\star_G$ and $\star_H$ are binary operations on $G$ and $H$ respectively.

In here we have, $$\left([b], \tau\right)^{-1}\left([a], \alpha\right)\left([b], \tau\right)=\left([b]^{-1}\star_1 [a] \star_1 [b], \,\,\, \tau^{-1}\alpha \tau\right).$$

Now observe that, $\mathbb{Z}_4$ is an abelian group, so $bab^{-1}=a$.

Also in $S_4$, $\tau (abc) \tau^{-1}=(\tau(a) \, \tau(b) \, \tau(c))$

This means
\begin{align*} \left([2], (123)\right)^{-1}\left([1], (24)\right)\left([2], (123)\right) &=\left([1], (321)(24)(123)\right)\\ &=\left([1], (14)\right) \end{align*}