Proof involving group of permutations of {1,2,3,4}.

[pondzo]

Homework Statement

Let $\sigma_4$ denote the group of permutations of $\{1,2,3,4\}$ and consider the following elements in $\sigma_4$:

$x=\bigg(\begin{matrix}1&&2&&3&&4\\2&&1&&4&&3\end{matrix}\bigg);~~~~~~~~~y=\bigg(\begin{matrix}1&&2&&3&&4\\3&&4&&1&&2\end{matrix}\bigg)$
$\sigma=\bigg(\begin{matrix}1&&2&&3&&4\\2&&3&&1&&4\end{matrix}\bigg);~~~~~~~~~\tau=\bigg(\begin{matrix}1&&2&&3&&4\\2&&1&&3&&4\end{matrix}\bigg)$

and put $K=\{1,x,y,xy\},~~~~~~~ Q=\{1,\sigma,\sigma^2,\tau,\sigma \tau,\sigma^2\tau\}$

Show that if $q\in Q$ and $k\in K$ then $qkq^{-1}\in K$.

Homework Equations

I have shown $K$ and $Q$ are subgroups of $\sigma_4$ that $\sigma_4=KQ=\{kq~~;~~k\in K,~q\in Q\}$.

And I have found the following relations: $x^2=1,y^2=1,yx=xy;~~\sigma^3=1,\tau^2=1,\tau\sigma=\sigma^2\tau$

The Attempt at a Solution

I would go through and calculate $qkq^{-1}$ for each $k$ and each $q$ but I know there must be a shorter way. Do you know of it?

 

[Orodruin]

Just check the relation for the (sub)group generators. You do not need to check every element.

 

[pondzo]

Ok thanks. So I only need to calculate $\sigma x (\sigma)^{-1},~\tau x (\tau)^{-1}, \sigma y (\sigma)^{-1} \text{ and } \tau y (\tau)^{-1}$?

 

[Orodruin]

Right. You should however think about and convince yourself why this is sufficient as well.

 

[Orodruin]

Also a personal preference: I would suggest you learn and start using the more short-hand notation for the symmetric groups in terms of cycles. In this notation, your elements would be
$$
x = (12)(34), \ y = (13)(24), \ \sigma = (123), \ \tau = (12).
$$
It really simplifies performing the group multiplications as well.