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Theorem 5.1.1 in Derek J. S. Robinson's book "A Course in the Theory of Groups" says

The class of soluble groups is closed with respect to the formation of subgroups, images, and extensions of its members.

I think this means that subgroups $H$ and quotients $G/N$ are solvable if $G$ is solvable. I don't know what the author means by "extensions of its members". Can someone please explain?

Shaun
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John Smith
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1 Answers1

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A group $G$ is called an extension of a group $Q$ by a group $N$, if $N$ is a normal subgroup of $G$ with quotient $G/N\cong Q$. In other words, we have a short exact sequence $$ 1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1. $$ Now $G$ is solvable if and only if $N$ and $G/N$ are solvable. So extensions of solvable groups are solvable. In other words, the class of solvable groups is closed under extensions of its members.

Note that the class of nilpotent groups is not closed under extensions.

Dietrich Burde
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  • Thank you. I presume by the word "member", the author means the normal subgroup N? Also, out of curiosity, can you provide an example that shows that the class of nilpotent groups isn't closed under extensions? – John Smith Jul 29 '21 at 19:37
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    @H.H. If $N$ and $Q$ are members of the class of soluble groups (i..e, are soluble groups), then so is $G$ in the above – Hagen von Eitzen Jul 29 '21 at 19:40
  • Though that doesn't matter here, note that there is actually ambiguity in "extension of $Q$ by $N$". It can either mean what you write, or the exact opposite! (that $Q$ is normal and $G/Q\cong N$). See the discussion here. You may have to/want to dig through the comments in my answer, too – Arturo Magidin Jul 29 '21 at 20:34
  • @H.H. $S_3$ is an extension of a cyclic group by a cyclic group, but is not nilpotent. – verret Jul 30 '21 at 01:27