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There is a famous lemma saying: "Every subnormal subgroup of a semisimple group is normal and is a semisimple direct factor of the group". Any hints about it? :)

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    by induction: it's true for normal subgroups, and those normal subgroups are again semisimple2011-04-28
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    @user8268: The induction argument will still require the defect 2 case to be done explicitly: if $H\triangleleft M\triangleleft G$, you need to show $H\triangleleft G$. Once you have this, then the induction argument works, but this does not follow inductively from the fact that normal subgroups of semisimple groups are normal.2011-04-28
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    @Arturo: Is that true for a semisimple group that it is centerless and perfect? Will these two facts help me to probe the lemma, regarding your comments, properly? Thanks :)2011-04-28
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    I was only pointing out that user8268's suggestion is incomplete, since one *really* needs to prove the defect 2 case rather than the normal subgroup case. If you start with an arbitrary subnormal subgroup $H$, take a subnormal series, and let $M$ be the penultimate term, then induction tells you that $H$ is normal **in M.** But you want to prove that $H$ Is normal in $G$, not just in $M$. And to conclude this, you need the defect at most 2 case.2011-04-28
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    user8268's proof can sort of work, but it has to be phrased carefully. Not carefully phrased it sounds like all you get are subnormal subgroups are semisimple, without getting normality. Arturo's suggestions also gives normality. However, you can carefully handle user8268's proof to give also *direct factor*, and direct factors of direct factors are direct factors (and normal), so this sort of induction does work.2011-04-28
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    Does $G$ semisimple mean that $G$ is the central product of subgroups $Q_1,..,Q_n \leq G$ such that $[Q_i,Q_i]=Q_i$ and $Q_i/Z(Q_i)$ is simple?2015-11-19

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