I found this proof. But among my colleagues it is rumored that there is a mistake. Could you tell me if I did, what would the mistake be? I really don't have much idea how this proof works.
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3It looks correct, but also way more complicated that necessary. – Tobias Kildetoft Nov 20 '20 at 19:30
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Alternatively, show that $[G\times H,G\times H]=[G,G]\times[H,H]$ and use induction on the number of factors. – Arturo Magidin Nov 20 '20 at 22:25
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Please don't rely on pictures of text – Shaun May 14 '22 at 11:25
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where $H_{11}$ is the smallest nontrivial subgroup of $G_1$ in a composition series...
That makes sense only when each $G_i$ is finite. Infinite solvable groups don't even have composition series: A solvable group with composition series is finite and don't have to have minimal subgroups.
The general solution would be due to:
Lemma. Let $G$ be a group and $H\subseteq G$ a normal subgroup. Then $G$ is solvable if and only if both $H$ and $G/H$ are.
For a proof see this: For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable?
With that we automatically obtain that $G\times H$ is solvable if and only if both $G$ and $H$ are. And then we apply a simple induction to conclude that $G_1\times\cdots\times G_n$ is solvable if and only if each $G_i$ is.
freakish
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