When exactly can you turn a group into a ring?

Question

Obviously when you can add a second operation satisfying the axioms, but is there an easy characterization of that?

What I mean, let's say I have an abelian group (G,+). When is it possible to keep its structure as a group and define a second operation to have a ring (G,+,×)?

What I want is basically for these to hold:

1. (ab)c=a(bc)
2. a(b+c)=ab+ac

and possibly
3. There exists element e such that for any x in G, ex=xe=x

Ok, how do I actually check these? I mean what I actually want to show is that such an operation is possible, it doesn't necessarily need to "feel like multiplication", so something like 3(2+1) must distribute as 3*2+3*1, but it could equal 17 for all we know. Does the operation even need to be surjective? We don't require inverses so maybe not.

This is just a weird though, I don't know how to approach this, but it just seems like there are some obvious conditions I will feel stupid for not noticing intuitively.

Answer 1

I have two ideas:

1. Trivial multiplication.  If you make g×h = 0 for all elements g,h of G, it will satisfy the axioms.

"it doesn't necessarily need to "feel like multiplication""

It sure doesn't.  But it gets the job done.

2. If you know your group is a subgroup of another ring's additive group, then you can check for closure of multiplication to see if your group is a subring.

That is to say, if your group G is a subgroup of (R, +) for some ring R, all you have to do is check that any two elements of G, multiplied, gives you an element of G.

Answer 2

To clarify, I wrote (G,+) but the group doesn't have to be under any kind of addition, I just wrote that because + is usually how the "first" ring operation is denoted. I'm looking for a general result, if it's a group that's not isomorphic to anything additive, that's still under consideration.

Answer 3

Unhelpful, but if your Abelian group is finite, it's a product of cyclical groups which might help answer your question

Answer 4

Every f.g. abelian group can do that. This is because it's going to be of the form Z^n x product of F_q for prime powers q, so you just put in the obvious product.

I think if it's not f.g. you should be able to pull off some Zorn's lemma argument, but I haven't thought it through.

Answer 5

If f,g are  group endomorphisms  and u is an element of the  **abelian** group then let “+”  be (f+g)(u) = f(u) +g(u)  and “*” be function composition (f*g)(u) = f(g(u)).

Since f and g are  both endomorphisms   f(g(u+v)) = f(g(u) + g(v))  = f(g(u)) + f(g(v)) fulfilling the compatibility of both operations that we usually call the distributive property.  You can verify this will always form a ring with identity.  This isn’t quite the same as just adding an operation to an existing group but think of it as a kinda of ring you get for “free”  from an abelian group.