Reading around I found the following two assertion:
1) Every countable abelian group has a recursively enumerable set of defining relations.
2) Every countable locally finite group has a recursively enumerable set of defining relations.
How can we prove this directly?
We need to find an algorithm which halts exactly when we give as input a relation of the group. As concern abelian groups, I was thinking in this way:
If $w$ is an element of the free group of countable rank, then:
- If $w$ is a commutator (I think we can check this in a finite number of steps), halt.
- If $w$ is not a commutator, then?
And for countable locally finite groups?
I don't believe these claims. Notice that, for any set $X$ of prime numbers, we can build a countable, abelian, locally finite group in which $X$ is the set of primes that occur as orders of elements. Just take the direct sum (not product, because I want it to be countable and locally finite) of cyclic groups $\mathbb Z/p$ for all $p\in X$. By varying $X$, we get uncountably many non-isomorphic groups of this sort. But there are only countably many recursively enumerable sets, so I don't see how you could have r.e. sets of relations to define all these groups.