# Thread: Algebra, Problems For Fun (31)

1. ## Algebra, Problems For Fun (31)

True or False: There exists a group $G$ such that $(\mathbb{Q},+) \cong \text{Aut}(G).$

2. I've been struggling with this one for a couple of hours.

The group (R,+) has (R-{0},*) as its automorphism group. The positive reals are a subgroup of index 2 isomorphic to (R,+) via the mapping x->log(x). The same mapping shows that there is a subgroup (R-{0},*) isomorphic to (Q,+). This shows that it is not completely implausible that a group G might have Aut(G) isomorphic to (Q,+)

On the other hand there are plenty of things which make it look like it is probably False.
Clearly G must be infinite since otherwise Aut(G) is finite.
The mapping g->g^-1 is an automorphism of order 2 if G is abelian unless every element of G has order 2. Since (Q,+) is torsion free then either G is abelian and every element has order 2, or G is non-abelian.
In the former case G cannot be finitely generated so we would expect Aut(G) to be isomorphic to a subgroup of the invertible linear transformations of an infinite dimensional vector space over F2. It is not easy to see how such a subgroup could possibly be isomorphic to (Q,+) : none of the transformations that fix all but finitely many generators can be in the subgroup, since these all have finite order.
In the latter case (G not abelian) there are distinct elements a,A such that g->A*g*a is a non-trivial automorphism (so a,A are inverses of each other). All such elements must be torsion-free since otherwise there would be an automorphism with finite order. Now the requirement that Aut(G) is isomorphic to (Q,+) means that for all (a,b,g in G) we have B*A*g*a*b=A*B*g*b*a (again using case change for inverse pairs of elements).
So G is probably not finitely presented even if it were finitely generated. Also with so many torsion-free elements one would expect to be able to find an automorphism that permuted finitely many generators and thus had finite order.

3. $(\mathbb{Q},+)$ is locally cyclic, i.e. for any $a,b \in \mathbb{Q},$ the subgroup $$ is cyclic. see if you can get something out of this.

4. Unfortunately I cannot, though I feel I nearly can.
If G is non-abelian with a*b <> b*a then the two mappings g->A*g*a and g->B*g*b are both non trivial member of Aut(G) . Now we should be able to find some other group element k in G such that the mapping g->K*g*k generates the subgroup of Aut(G) generated by these two automorphisms (since Inn(G) is a subgroup of Aut(G)). Now the orbits of both a and b under this mapping are both finite since for some m,n in N we have K^m*a*k^m = a and K^n*b*k^n=b. On the other hand K*a*k and K*b*K are distinct from a and b respectively because otherwise the mapping could not possibly generate the subgroup.
I'd got this far because I knew any finitely generated subgroup of (Q,+) was cyclic, though I'd forgotten the term locally cyclic if I ever knew it, but I can't see how to get much further. Given any finite set of elements S from outside the centre of G we can always find an element K of G and a natural number n such that K^n*s*k^n =s and K*s*k <> s for all s in S. I'd like to say that this shows that Aut(G) has a torsion element which would contradict Aut(G) being isomorphic to (Q,+) but I haven't shown that K^n*g*k^n=g for the whole of G, though this shows that G cannot be finitely generated, since if it were Inn(G) would be finitely generated, hence cyclic and we could find a torsion element this way. But I can't rule out G being non-Abelian torsion free and not finitely generated. Such a G would be very intractable - for example every generator would have to occur somewhere in a presentation, otherwise G would have a free subgroup and then there would be an automorphism that interchanged a finite number of generators and hence was a torsion element of G.