Groups with finite automorphism groups.
An easy argument shows that for any finite group $G$ the cardinal of $Aut(G)$ is less than $(|G|-1)!$. In particular the automorphisms group of a finite group is finite. Basically my question is about the converse statement. If a group $G$ has finite automorphisms group then should $G$ be finite ?
The answer to this question is no, the example of $\mathbb{Z}$ with $Aut(\mathbb{Z})$ isomorphic to $\mathbb{Z}/2\mathbb{Z}$ is a counter-example.
Hence, using this, it is an easy exercise to show that for any finite group $F$ we must have $Aut(\mathbb{Z}\times F)$ finite of cardinal less than $2\times |F|\times Aut(F)$ (basically, $F$ is a characteristic subgroup of $\mathbb{Z}\times F$ hence $(1,1_F)$ must be sent to $(\pm 1,f)$ where $f\in F$) leading to a family of counter-examples.
Now take a group $G$ with finite automorphisms group then $G/Z(G)$ is finite. Hence any such $G$ can be written as a central extension :
$$\text{ (E) : }1\rightarrow A\rightarrow G\rightarrow F\rightarrow 1 $$
Where $A$ is abelian and $F$ finite. Now come the real questions (remarks, critics, partial answers and references are welcome) :
- Can we have a characterization of groups $G$ with $G$ infinite abelian and $Aut(G)$ finite ?
I believe that such groups are isomorphic to $\mathbb{Z}\times F$ where $F$ is a finite abelian group. It is certainly true when we assume $G$ to be finitely generated and as for non-finitely generated groups, I know for a fact that $\mathbb{Q}$ will have $\mathbb{Q}^*$ as automorphism group. Although I am not totally sure of the argument $\mathbb{Z}(p^{\infty})$ (the set of roots of unity of order $p^k$ for some $k$) also has infinite automorphisms group. As for "general" non-finitely generated abelian groups I have no clue.
- Can every group $G$, infinite group with finite automorphisms group, be written as a central extension $(E)$ with $F$ finite, $A$ abelian with finite automorphisms group?
Clearly if 1 admits a precise answer and 2 a positive answer then we have a good way to reduce the problem. The last question is the converse.
- Assume that a group $G$ is written as a central extension $(E)$ where $F$ is finite and $A$ infinite abelian group with finite automorphisms group then does it follow that $G$ is infinite with finite automorphisms group?
Please note that the extension needs to be central, in particular $G:=\mathbb{Z}\rtimes_{-1}\{\pm 1\}$ (the fundamental group of the Klein bottle) is not a counter example to 3.
Ok thanks to the reference given in the comment, I can give the answer to those questions.
First a corollary from a result in " V. T. NAGREBECMI. On the periodic part of a group with finitely many automorphisms. Doll. Akad. Nauk. SSSR 205 (1972). 519-521: Sol,ier Math. DoXI. 13 (1972). 953-956."
If $G$ is an infinite abelian group with $|Aut(G)|<\infty$ then there exists a subgroup $F$ without torsion of $G$ such that $G=Tor(G)\oplus F$ and $Tor(G)$ is a finite group.
Hence we see that to answer question 1 it suffices to deal with infinite abelian groups without torsion, but with the following citation taken in J. T. HALLETT AND K. T. HIRSCH. Torsion-free groups having finite automorphism groups, I. J. Algebra 2 (1965). 287-298., we see that it is hard :
"we do not concern ourselves with the apparently hopeless task of finding all the torsion free abelian groups whose automorphism group is a given finite group"
Following their comment, we can say that 1. is "hopeless".
However question 3 admits an answer. The answer is yes if $A$ is the center of $A$. The proof goes as follow, any $\phi\in Aut(G)$ induces an automorphism of $\varphi\in Aut(Z(G))$ and an automorphism $\psi\in Aut(Q)$ if $\iota$ is the inclusion of $Z(G)$ in $G$ and $\pi$ the projection of $G$ onto $F$ the automorphisms verify :
$$\iota\circ\varphi=\phi\circ\iota\text{ and } \pi\circ \phi=\psi\circ \pi $$
On the whole this gives a group morphism :
$$\mathcal{A}: Aut(G)\rightarrow Aut(Z(G))\times Aut(F) $$
Furthermore its kernel $Ker(\mathcal{A})$ is easily seen to be isomorphic to $Hom(F,Z(G))=Hom(F^{ab},Z(G))=Hom(F^{ab},Tor(Z(G)))$.
Assume that $Aut(Z(G))$ is finite then by the result above $Tor(Z(G))$ is finite hence $Ker(\mathcal{A})$ is finite and because $Im(\mathcal{A})$ is in a finite subgroup, it must be finite as well, on the whole $Aut(G)$ is finite.
Now for the question 2, I did not find a counter-example but it should be false.