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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 $\sigma_4=KQ=\{kq~~;~~k\in K,~q\in Q\}$.

There must be a shorter method than explicitly multiplying out each of the elements of K and Q to show that each corresponds to one of the $24$ elements in $\sigma_4$. Although I can't see what it is, maybe one of you know?

I have found the following relations: $x^2=1,y^2=1,yx=xy;~~\sigma^3=1,\tau^2=1,\tau\sigma=\sigma^2\tau$ I have also shown that K and Q are subgroups of $\sigma_4$.

NormalsNotFar
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  • I think there is a typo with the definition of $x$. Also, it is more standard to denote the symmetric group by $S_n$, rather than $\sigma_n$. – Batominovski Feb 10 '16 at 17:18
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    If you use the right $x$, then $K$ will be a normal subgroup of $G$ such that $K\cap Q$ is trivial. So, $KQ$ is a subgroup of $G$, but as $|KQ|=|K|,|Q|=24=|G|$ (due to the Second Isomorphism Theorem $KQ/K \cong Q/(K\cap Q)$), $G=KQ$. – Batominovski Feb 10 '16 at 17:20
  • Hi Batonminovski, how did you come to the conclusion that K was normal in G ? Did you Take a look at the left and right cosets? – NormalsNotFar Feb 10 '16 at 19:39
  • Your $K$ is definitely not normal in $G$. It isn't even a subgroup and $KQ\neq G$. So, unless you fix the definition of $K$, there is no point in answering the question. (Observe that $x=\tau$.) – Batominovski Feb 10 '16 at 22:28
  • Sorry, i know there was a mistake in the definition of x, it was a typo. Could you answer with the correction to the post? – NormalsNotFar Feb 10 '16 at 22:41
  • The question about normality of $K$ in $G$ has been answered here before. See for example here: http://math.stackexchange.com/questions/1370950/prove-that-the-4-group-v-is-normal-subgroup-of-s-4-by-using-isomorphism-theore/1370973#1370973. – Batominovski Feb 10 '16 at 23:29

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