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I know that the direct product of two solvable groups are solvable. The group $S_3$ is solvable, so $S_3\times S_3$ is solvable. But how am I going to establish the subnormal series of $S_3\times S_3$?or is there any simpler way to show its solvability?

Thanks.

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3 Answers 3

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If you've got a composition series $1 \trianglelefteq H \trianglelefteq S_3$ (I hope you do!), then you can use that to refine $1\times 1 \trianglelefteq 1 \times S_3 \trianglelefteq S_3 \times S_3$ into a series where the factors are abelian.

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Perhaps this is one of those cases where you understand things better by looking at a more general setting.

Let $G, H$ be soluble groups, and let $G.H$ be any extension of $G$ by $H$. Then $G.H$ is soluble.

Start with a subnormal series with abelian factors that goes from $\{1\}$ to $G$. Then continue with a subnormal series with abelian factors of $H$, or to be precise, with the counterimages of the elements of such a series through the epimorphism $G.H \to H$.

In your case $G.H = G \times H$. So you simply start with the required subnormal series for $G$, and then from $G$ you continue with $G N$, with $N$ in the required subnormal series for $H$.

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$$\langle\omega\times\omega\rangle \trianglelefteq \langle\omega\rangle\times A_3 \trianglelefteq A_3\times A_3 \trianglelefteq A_3\times S_3 \trianglelefteq S_3\times S_3$$

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  • $\begingroup$ What is $\omega$? $\endgroup$
    – martini
    Jul 14, 2014 at 8:45
  • $\begingroup$ I mean the identity permutation $\endgroup$
    – Bumblebee
    Jul 14, 2014 at 8:49

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