Groups with given automorphism groups

Question

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 $\text{Aut}(G)=1$ (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 $\text{Aut}(G)\cong A$?

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?

Answer 1

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).

Answer 2

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 $>1$.]

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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).

Answer 3

A more accessible account on the theorem of Ledermann-Neumann mentioned in the accepted answer can be found here: Math. Monthly or arxiv

EDIT: The argument runs as follows. Suppose $G$ is a finite group with exactly $n$ automorphisms.

1. The order of the derived group $G'$ is bounded in terms of $n$. Since the size of the inner automorphism group $|G/Z(G)|$ is bounded by $n$, this follows from a theorem of Schur. While Schur's proof is highly non-trivial, there is an elementary argument by Rosenlicht.
2. Every prime divisor of $|G|$ is bounded by $n+1$. This requires a special case of the Schur-Zassenhaus theorem for central Sylow subgroups.
3. The exponent of $G$ is bounded in terms of $n$. This uses an elementary argument by Nagrebeckiı̆.
4. The size of the center $Z(G)$ is bounded in terms of $n$. Since the exponent is bounded, it suffices to bound the minimal number of generators of $Z(G)$. Let $g_1,\ldots,g_m$ representatives for the cosets of $G$ mod $Z(G)G'$ and consider $U:=\langle g_1,\ldots,g_m\rangle G'$. Note that $m$ is bounded in terms of $n$. Using another lemma, one finds a decomposition $Z(G)=C\times D$ such that $U\cap Z(G)\le C$ and $|C|$ is bounded. It remains to bound $|D|$. The author shows that $G=UC\times D$. Now we may assume that $G=D$ is an abelian $p$-group. It is easy to see that the number of automorphisms of $G$ grows with its rank.