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$A$ and $B$ are finite groups. $H \leq A \times B$. Can we find some $C \leq A$, $D \leq B$ such that $H \cong C \times D$?

In case the statement is not true: is it true under further assumptions about A and B, such as solvability, nilpotency, etc?

Special cases I can prove:

  1. $A$ and $B$ are abelian (following ideas from another discussion: $G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?)

  2. $(|A|,|B|)=1$. In this case we even have $H = C \times D$. By using the Chinese remainder theorem for instance.

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up vote 10 down vote accepted

No, a subgroup of a direct product need not be a direct product of subgroups, even up to isomorphism, even if the factors are almost abelian.

Examples are plentiful. Here is a small nearly-abelian example (nilpotent and solvable and hamiltonian) and a small easy to write down example:

  1. Take A=B to be quaternion of order 8. A×B has 15 maximal subgroups of order 32, 9 of which are directly indecomposable, and so no such C or D exist. Hence nilpotent of class at most 2 with all subgroups normal (just barely not-abelian) is not sufficient.

  2. Take A=B to be non-abelian of order 6. A×B has a maximal subgroup H of order 18 that is directly indecomposable, so again no such C or D exist. H can be generated by (1,2,3), (4,5,6) and (1,2)(4,5). Its elements of order 2 are all of the form (a,b)(c,d) where a≠b in {1,2,3} and c≠d in {4,5,6}, and so all of them are self-centralizing, and so none of them can be part of a proper direct factor (which is centralized by the entire other non-identity factor). One of the direct factors has to have even order, and so that one has to be all of H.

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Thank you Jack! The second example is indeed easy to understand and the first one is insightful. What is your method to count the subgroups (in the first example: 15 maximal subgroups, 9 of which..). Maybe I'll know more about the counting after reading the article Jonas Meyer referred to. – user3533 Dec 11 '10 at 22:57
If a p-group can be generated by k elements, but not by k-1, then it has exactly (p^k-1)/(p-1) maximal subgroups, each of index p, each normal. This is called the Burnside basis theorem. Q8xQ8 has k=4, and so 15 maximal subgroups. I didn't work out an easy way to count the two types of maximal subgroups (there probably is one, Goursat's lemma should work). I just asked GAP to examine them for me. They are a lot like the H in #2. In #2 there is just one H, but in #1 there are 9, all isomorphic and indistinguishable, so which one should I distinguish? – Jack Schmidt Dec 12 '10 at 2:42

You might be interested in the expository article "Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat's lemma" by Anderson and Camillo. A couple of excerpts:

alt text

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Thank you very much. – user3533 Dec 11 '10 at 22:59

It's not possible if you want the isomorphism to be compatible with the product structure.

Namely, choose $A=B$ and $H = \lbrace (a,a) : a\in A\rbrace$ the "diagonal subgroup". Clearly, the subgroup $H$ does not have the form $H = C\times D \subseteq A\times A$.

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Makes me wonder how exactly he "proved" the abelian case, since your argument works for an abelian group as well. – Raskolnikov Dec 11 '10 at 13:57
@Raskolnikov: I'm not requiring the isomorphism to be compatible with the product structure. That's why this argument does not contradict my proof of the abelian case. Note what I wrote about the case $(|A|,|B|)=1$ in which the isomorphism happens to be compatible with the group structure. – user3533 Dec 11 '10 at 13:59
Thank you for this answer. The diagonal subgroup was part of the motivation for my Q, because although it's not of $C \times D$ form, it is isomorphic to $A \times 1$. – user3533 Dec 11 '10 at 14:10
You are answering a different question: namely whether the subgroup lattice of the direct product is equal to the direct product of the subgroup lattices. – Alex B. Dec 11 '10 at 14:12

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