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What is the smallest set of groups $S$, such that for any group $G$ there exists either $H \in S$ or $H = H_1 \times H_2 \times \dots \times H_n$ for $H_i \in S$ such that $G$ and $H$ are isomorphic.

If anything interesting can be said about them, I'd like to hear all cases : with and without infinite groups, with and without infinite products.

Notes. I realize that there may not be a unique such set. My own only idea is the set of all non-abelian groups and cyclic groups of prime order. Also, it should be mentioned that by "small" I mean $\subset$ relation.

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Any simple group of your considered family must belong to $S$, for a direct product of two non-trivial groups has at least one normal subgroup, namely $G \times 1$ and $1 \times H$ are normal subgroups of $G \times H$. I think you also need to add those groups with precisely one normal subgroup in $S$, for $G \times H$ will always have at least two non-trivial normal subgroups... this would be a "lower bound" in the sense of partial order by inclusion, to your minimum. – Patrick Da Silva Aug 10 '12 at 7:23
In the case of finite groups, $S$ should be precisely those groups without any direct factors, i.e. the directly indecomposable groups. I wonder if the induction can be extended transfinitely for arbitrarily infinite groups. – anon Aug 10 '12 at 7:43
"Monoid of all groups"? You might want to read this... – user1729 Aug 10 '12 at 14:47
@anon: I think there are abelian groups that are neither directly indecomposable, nor a direct product of directly indecomposable groups (for instance a countably infinite dimensional vector space). If one uses direct sum (restricted direct product), then everything is ok for abelian groups, and probably groups in general. – Jack Schmidt Aug 10 '12 at 15:30
@JackSchmidt I was afraid something like that would happen, where every nontrivial decomposition is too weak to break down a group's size or structure. That makes the infinite case much harder (for products). Thanks. – anon Aug 10 '12 at 16:28

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