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The question I want to ask is actually slightly broader than that in the title: what is the smallest class of finite groups which contains all finite simple groups groups, and is closed under semidirect products?

This arose out of a conversation I was having with a friend a while back. At the time, I thought I saw a simple argument that this class of groups was in fact the class of all finite groups, but that argument turned out to be gibberish. (EDIT: As Arturo points out below, this class cannot possibly consist of all finite groups.)

Thanks in advance for the help!

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    I don't know the answer to this right now, but the smallest class that contains all finite simple groups and is closed under subgroups, quotients, and products (i.e., a variety) is the class of all groups.2012-04-02
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    Can't be the class of all finite groups: the only abelian groups that occur are groups that are products of elementary abelian $p$-groups (only groups that are products of finite abelian simple groups can occur).2012-04-02
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    Good point - thanks!2012-04-02
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    If a group in this class has odd order then it is necessarily a product of groups of prime order, because every group of odd order is soluble (by the Feit-Thompson theorem, and so $G$ simple and of odd order if and only if $G$ has prime order greater than $2$). You then know, for some value of "know", how these pin together as semidirect products, as $\operatorname{Aut}(C_p)=C_{p-1}$2012-04-02
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    @user1729: I would caution that "know" is used in a very weak sense. Every p-group is a subgroup of a p-group in this class of groups, even though the Sylow p-subgroup of Aut(C_p) is trivial (and so one would have imagined that p-groups in this class would be abelian, but this is very far from true). One has to also understand the automorphism groups of full groups in this class, and they can be quite complex (complex enough to contain any group as a subgroup).2012-04-02
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    I believe no group in this collection can have a cyclic Sylow 2-subgroup of order greater than 2.2012-04-02
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    @JackSchmidt: That's very interesting. I'd omitted to think that you'd need to know $\operatorname{Aut}(C_p\rtimes C_p)$, etc. Why is it that every $p$-group is a subgroup of a $p$-group in this class of groups?2012-04-03
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    @User1729: The Sylow p-subgroups of symmetric groups are in the collection, and every group is a subgroup of a symmetric group. The class is closed under direct products and then semi-direct products, so it is closed under wreath products, and the Sylow p-subgroup of a symmetric group is a direct product of iterated wreath products of cyclic groups of order p.2012-04-03

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