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Is the solvable radical of the commutator subgroup $G'$ of a group $G$ normal in $G$?

Ronald
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    Informally, you can apply what Martin Isaacs calls the "the" test: if you can use the definite article "the", such as "the commutator subgroup of $G$" and "the solvable radical of $G'$", then the subgroup in question is characteristic. Specifically, the solvable radical of $G'$ is characteristic in $G'$, and $G'$ is characteristic in $G$. Since "characteristic" is a transitive property, i.e. $A \operatorname{char} B \operatorname{char} C$ implies $A \operatorname{char} C$, it follows that your solvable radical is in fact characteristic in $G$. –  Feb 26 '18 at 20:34
  • That's nice thanks :) – Ronald Feb 26 '18 at 20:37

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Yes, because the commutator subgroup is a characteristic subgroup of $G$. Then the claim follows from this duplicate:

Prove that if $H$ is a characteristic subgroup of $K$, and $K$ is a normal subgroup of $G$, then $H$ is a normal subgroup of $G$

Dietrich Burde
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