all 10 comments

[–]Ashtero 13 points14 points  (1 child)

Yes, it is possible!

For example, take free groups on 2 and 4 generators F2 and F4. Obviously F2 is a subgroup of F4. They are not isomorphic -- for example, they have different amount of homomorphisms to Z/2Z. As for injection of F4 into F2, here is one example:
a -> x;
b -> yxy^-1;
c -> y^2xy^-2;
d-> y^3xy^-3.

Edit: I interpreted your question as "is it possible for g to be isomorphic to a subgroup of H and vise versa". If you literally meant that they should be actual subgroups of each other, but not isomorphic, then it is indeed obviously not possible.

[–]deaths_accountant[S] 8 points9 points  (0 children)

That is what I meant. Thank you for answering what I meant rather than what I said.

[–]WibbleTeeFlibbet 1 point2 points  (4 children)

It's not possible for non-isomorphic groups.

If g is a subgroup of H, there's an inclusion map g -> H given by g |-> g. And since H is a subgroup of g, there's an inclusion H -> g defined by h |-> h.

We can show that both of these inclusions are in fact isomorphisms. That they're injective homomorphisms is obvious. To see they're also surjective, let h in H and consider the inclusion of h into g. This is an element h in g. Now consider the inclusion of h in g into H. This is an element of g that maps to h in H, so the inclusion g -> H is surjective. Similarly for the inclusion H -> g.

Thus if g and H are subgroups of each other, they are isomorphic.

[–]deaths_accountant[S] 1 point2 points  (0 children)

Sorry, this is because I worded it imprecisely. I really meant that g is isomorphic to a subgroup of H, and vice versa, so the inclusion map didn't work. And u/Ashtero showed that it is possible, as long as I don't mess up my wording of the question.

[–]gnome_genome 0 points1 point  (2 children)

If g and H were identical, wouldn't they both be included in each other?

[–]WibbleTeeFlibbet 1 point2 points  (1 child)

Yes, but the extra text of the OP says g and H aren't isomorphic. I updated my answer to clarify.