This is a sort of catch-all section for all the other results I have relating to dominions in the setting of categories of groups. First let me address another specific class of varieties, and then I will indulge in a bit of more abstractness.

Simple Groups

   As already mentioned, P.M. Neumann proved that no category of solvable groups which is closed under quotients can have nonsurjective epimorphisms. However, if we start with the alternating group on five letters A(5), then B.H. Neumann proved that in the variety it generates (the collection of all homomorphic images of subgroups of powers of A(5)), the embedding of A(4) into A(5) is an epimorphism, and trivially nonsurjective. See: Splitting groups and projectives in varieties of groups by P.M. Neumann, Quart. J. Math. Oxford (2), 18 (1967), 325-332.

   By generalizing B.H. Neumann's argument, one can describe the dominion of a subgroup H of a finite nonabelian simple group S in the variety generated by S. Namely, it is the subgroup fixed by all automorphisms of S which fix H pointwise.

   This in turn generalizes to any variety generated by a family of finite nonabelian simple groups. It turns out that any such variety is either generated by only finitely many such groups, or else is the variety of all groups. In the latter case, dominions are of course trivial. In the former, if one picks a minimal such family, then the result above still holds for subgroups of the generating groups.

   As a consequence of this, there are plenty of varieties with nonsurjective epimorphisms. Let us define a subgroup to be totally variant if the only automorphism of the group which fixes the subgroup pointwise is the identity. Then any totally variant proper subgroup is epimorphically embedded into a finite nonabelian simple group in the variety it generates. It is easy to show that A(n) is totally variant in A(n+1), n>3, so this gives examples. I have also been assured that every finite nonabelian simple group contains a proper totally variant subgroup.

Dominions in product varieties

   If we have two varieties V and W, we can form their product VW, which consists of all groups which are extensions of a V group by a W group. A variety is decomposable if it can be written as such a product, where neither factor equals E (the trivial variety consisting only of the trivial group) or O (the total variety consisting of all groups). Otherwise it is indecomposable.

   By a theorem of B.H. Neumann, Hanna Neumann, P.M. Neumann, and G. Baumslag, the semigroup of varieties is a cancellation semigroup with 0 and neutral element, freely generated by the indecomposable varieties. See Varieties of Groups by Hanna Neumann.

   How much does the description of dominions in a product NQ depend on dominions in N and in Q? At first I thought it might completely depend on them but, as the example of metabelian groups shows, this is not the case.

   A few things can be said, though. Let G be an NQ-group, H a subgroup, and let Q(G) be the least normal subgroup of G such that Q(G) is an N-group and G/Q(G) a Q-group. Let J be the intersection of H and Q(G), and let D be the dominion of J in Q(G) in the variety N.

   The dominion of H in G in NQ will always be bounded below by <H,D> = HD, and above by HQ(G). It is not hard to show that it can fall at one extreme or somewhere in the middle in different circumstances. If we have one more bit of information, namely, if the normalizer M of D in G is "normal enough" (explicitly, if MQ(G)=G), then the dominion is exactly equal to HD.

   A consequence of the bounds is that if N is a variety in which there are nontrivial dominions, and Q is any variety other than the variety of all groups, then NQ necessarily has instances of nontrivial dominions (although they could be distinct from the ones we had in N; i.e. all those may collapse to triviality in the larger variety).

   Perhaps unfortunately it is not possible to describe dominions in a product variety completely in terms of the factors. Fortunately, in the case of nonsurjective epimorphisms, the result above is powerful enough to reduce the problem to the indecomposable varieties. Explicitly, in the situation and notation above, then the following are equivalent:

• H is epimorphically embedded into G in NQ.
• HQ(G)=G and D=Q(G).
• HQ(G)=G and J is epimorphically embedded into Q(G) in N.
• For every normal subgroup K of G with K in N and G/K in Q, HK=G and the intersection of H and K is epimorphically embedded into K in the variety N.

   In particular, if a product variety has nonsurjective epimorphisms, then so does the first fact. (This is also a consequence of McKay's argument in Surjective epimorphisms in classes of groups, S. McKay, Quart. J. Math. Oxford (2) 20 (1969), 87-90).

   The converse to that is still open. I do have two partial results in that direction: if the epimorphism is into a finite nonabelian simple group G, then for any Q other than the variety of all groups NQ will have a nonsurjective epimorphism into a finite group. If we only assume that G is finite, then NQ will also have a nonsurjective epimorphism provided that Q is a product of nilpotent varieties of exponent 0.