It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).

Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?

If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?

And if not, which groups arise as counterexamples?

And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?

  • $\begingroup$ If you talk about outer automorphism groups, so $\operatorname{Out}(G)=\operatorname{Aut}(G)/\operatorname{Inn}(G)$, and allow the groups $G$ to be infinite then for every countable group $Q$ there are infinitely many groups $G$ such that $\operatorname{Out}(G)\cong Q$. See, for example, this paper: arxiv.org/abs/1709.06441 (there are other similar results, but ensuring there are infinitely many non-isomorphic groups $G$ is easy here as the groups are HNN-extensions of triangle groups $\langle x, y; x^i, y^i, (xy)^i\rangle$, so changing $i$ immediately changes your group $G$.) $\endgroup$ – user1729 Nov 9 '17 at 17:14

Ledermann and B.H.Neumann ("On the Order of the Automorphism Group of a Finite Group. I", Proc. Royal Soc. A, 1956) have shown the following:

Theorem. Let $n > 0$. There exists a bound $f(n)$ such that if $G$ is a finite group with $|G| \geq f(n)$, then $|\operatorname{Aut}(G)| \geq n$.

An immediate consequence is that up to isomorphism, there are only finitely many finite groups $G$ with $|\operatorname{Aut}(G)| \leq n$. Hence for any finite group $X$, up to isomorphism there are only finitely many finite groups $G$ with $\operatorname{Aut}(G) \cong X$.

Among infinite groups this is no longer true, and indeed there are infinitely many groups $G$ with $\operatorname{Aut}(G) \cong \mathbb{Z} / 2 \mathbb{Z}$.

Then there is of course the question of determining all finite groups $G$ with given automorphism group $\operatorname{Aut}(G) \cong X$. For this, see for example

Iyer, Hariharan K. On solving the equation Aut(X)=G. Rocky Mountain J. Math. 9 (1979), no. 4, 653–670.

This paper gives a solution to the problem in some cases, and determines for example all $G$ with $\operatorname{Aut}(G) \cong S_n$. There is also a different proof of the fact that there are only finitely many groups with a given automorphism group (Theorem 3.1 there).

  • $\begingroup$ Is there any work on an explicit form for $f(n)$? $\endgroup$ – JamalS Nov 9 '17 at 22:17
  • $\begingroup$ @Mikko: Great answer; thanks a lot, also for the references (the Iyer paper contains some very interesting results)! $\endgroup$ – jpvee Nov 10 '17 at 6:26
  • $\begingroup$ @JamaS: Ledermann-Neumann give an explicit $f(n)$, something around $(n-1)^{2n}$. I guess there must be some further improvements in the literature. One result is that if $G$ is a nontrivial finite abelian group, then $|\operatorname{Aut}(G)| \geq \phi(|G|)$ with equality iff $G$ is cyclic. $\endgroup$ – Mikko Korhonen Nov 10 '17 at 8:57

Mikko's nice answer concerns finite groups $G$. Let me here answer for infinite groups $G$ (but still finite automorphism groups, as in the question).

The picture is indeed very different:

For $A=C_2$ cyclic, there exists uncountably many non-isomorphic (abelian countable) groups $G$ with $\mathrm{Aut}(G)\simeq C_2$.

Indeed, for $I$ a set of primes, let $B_I$ be the additive subgroup of $\mathbf{Q}$ generated by $\{1/p:p\in I\}$. Then $B_I$ and $B_J$ are isomorphic if and only if the symmetric difference $I\triangle J$ is finite, and $\mathrm{Aut}(B_I)=\{1,-1\}$ (easy exercise: more generally for a nonzero subgroup $B$ of $\mathbf{Q}$, its automorphism group is $\{t\in\mathbf{Q}^*:tB=B\}$ acting by multiplication).

One also gets the group $C_2^n$ ($n\ge 1$) in a similar fashion. Say, for $n=2$, choose $I,J$ such that both $I\smallsetminus J$ and $J\smallsetminus I$ are infinite: then $\mathrm{Aut}(B_I\times B_J)\simeq C_2\times C_2$.

In general, if a group $G$ has finite automorphism group $A$, then its center has finite index in $G$, because $G/Z(G)$ embeds into $A$. A well-known result then implies that $[G,G]$ is finite.

[Also, it follows that if $A$ is cyclic of odd order, we deduce that $G/Z(G)$ is cyclic, and hence $G$ is abelian, and then $G$ has to be a finite elementary abelian $2$-group, and then $G=1$ or $G\simeq C_2$, whence $A=1$. In other words, for no group (finite or infinite) $G$, $\mathrm{Aut}(G)$ is cyclic of odd order $>1$.]

One more example to mention that one gets non-abelian groups: let $F$ be a finite group. Then for every torsion-free abelian group $B$, $\mathrm{Aut}(B\times F)$ is a semidirect product $(\mathrm{Aut}(F)\times\mathrm{Aut}(B))\ltimes\mathrm{Hom}(B,Z(F))$. If $\mathrm{Aut}(B)=\{\pm 1\}$, then the $\mathrm{Aut}(B)$-action on $\mathrm{Hom}(B,Z(F))$ is trivial and this reduces to the product $\mathrm{Aut}(B\times F)=(\mathrm{Aut}(F))\ltimes\mathrm{Hom}(B,Z(F))\times\mathrm{Aut}(B)$. For $B=B_I$, we have $\mathrm{Hom}(B_I,Z(F))\simeq Z(F)$. For instance, for $F=C_2$ one gets $\mathrm{Aut}(B_I\times C_2)\simeq C_2^2$. The smallest non-abelian group we can get this way has order 12, namely for $F=C_3$ or $F=D_6$ (dihedral group of order 6), one gets $\mathrm{Aut}(B_I\times F)\simeq D_6\times C_2$. For $F=C_4$ one gets $\mathrm{Aut}(B_I\times C_4)\simeq D_8\times C_2$.

I don't know if we can obtain abelian $\mathrm{Aut}(B_I\times F)$ when $|F|\ge 3$. This holds if and only if $\mathrm{Aut}(F)$ is abelian and acts trivially on $F$. Then $F$ is non-abelian, of nilpotency class 2. Possibly some large $p$-groups satisfy this (see Jain-Rai-Yadav (arXiv link) for a discussion of large $p$-groups with abelian automorphism groups; however they don't indicate if they can be chosen so that automorphisms are trivial on the center).

  • $\begingroup$ Btw, out of curiosity: I could't determine if there exists an infinite group $G$ with $|\mathrm{Aut}(G)|=6$. $\endgroup$ – YCor Nov 10 '17 at 14:22
  • 2
    $\begingroup$ See "de Vries, H.; de Miranda, A. B. Groups with a small number of automorphisms. Math. Z 68 1958 450–464.". It follows from results in this paper that there exists an infinite abelian group with $\operatorname{Aut}(G) \cong C_6$ (pg. 456), but there does not exist any infinite group with $\operatorname{Aut}(G) \cong S_3$. $\endgroup$ – Mikko Korhonen Nov 10 '17 at 15:08

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