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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.2012-08-10
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    In the case of finite groups, $S$ should be precisely those groups without any direct factors, i.e. the [directly indecomposable groups](http://groupprops.subwiki.org/wiki/Directly_indecomposable_group). I wonder if the induction can be extended transfinitely for arbitrarily infinite groups.2012-08-10
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    "Monoid of all groups"? You might want to read [this](http://math.stackexchange.com/questions/172743/group-of-groups)...2012-08-10
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    @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.2012-08-10
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    @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.2012-08-10
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    @user1729, I don't see why one equivalence relation is more suited to define algebraic structures than another. Anyway, it seemed like a short way to name the problem.2012-08-10
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    @anon, what is known about directly indecomposable groups? There isn't a list, like for simple groups, is there? I don't see a lot of information about them on the net.2012-08-10
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    Yeah, there is no list of indecomposable groups. "Most" non-abelian groups are indecomposable, and there is no "nice" description of them. I describe how there are a zillion indecomposable groups in this http://math.stackexchange.com/questions/16781/is-there-a-classification-of-all-finite-indecomposable-p-groups2012-08-10

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