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I proved that the direct product of two solubles is soluble and I immediately had some questions in my head, is the product of a soluble with any group soluble? (If I were to kick, would I say no, some against example?)

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You're right, the answer is no. Let $S$ be a soluble group, and let $G = S\times A_5$, where $A_5$ is the (simple) alternating group of degree $5$. If $G$ were soluble, so would all of its subgroups be soluble. But, it has an insoluble subgroup isomorphic to $A_5$.

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  • $\begingroup$ Why not simply take $\{e\}\times G$ where $G$ is any non soluble group ? $\endgroup$ – Max Jun 26 '17 at 17:36
  • $\begingroup$ @Max Sure, that would work too. Anything of the form $S\times I$, with $S$ soluble and $I$ insoluble could form a counter-example. $\endgroup$ – James Jun 26 '17 at 17:43
  • $\begingroup$ It's just that it seemed simpler as $G\times\{e\} \simeq G$ :p but your answer is of course fine and provides a wider range of examples $\endgroup$ – Max Jun 26 '17 at 17:48
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You proved that the direct product of two soluble groups is again soluble. But the converse holds too. This is because any quotient of a soluble group is again soluble. Hence if $G_1 \times G_2$ is soluble then $G_1 \cong (G_1 \times G_2) / (\{e\} \times G_2)$ is soluble. Same for $G_2$.

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