Is every finite group the outer automorphism group of a finite group?

This question is inspired by this recent question, which essentially asks if we can realise every finite group $Q$ as the automorphism group of some group $G_Q$. The answer is well-known to be no, with the counter-examples being cyclic groups of odd order.

On the other hand, it is a theorem of Matumoto that every group $Q$ is the outer automorphism group of some group $G_Q$ [1]. It seems to be a research theme to place restrictions on the groups involved. For example, Bumagin and Wise proved that if we restrict $Q$ to be countable then we may take $G_Q$ to be finitely generated [2], and more recently Logan proved that if we restrict $Q$ to be finitely generated and residually finite group then we may take $G_Q$ to be residually finite [3, Corollary D] (this paper also cites quite a few other papers which play this game).

However, all the results I have found always produce infinite groups $G_Q$, even when the "input" groups $Q$ are finite. For example, Matumoto's groups $G_Q$ are fundamental groups of graphs of groups (so are always infinite), Bumagin and Wise use a variant of Rips' construction (so (as $Q$ is finite) their groups $G_Q$ have finite index in metric small cancellation groups, so are infinite), and Logan's groups $G_Q$ are HNN-extensions of hyperbolic triangle groups (so again are infinite). So we have a question:

Does every finite group $Q$ occur as the outer automorphism group of some finite group $G_Q$?

[1] Matumoto, Takao. "Any group is represented by an outerautomorphism group." Hiroshima Mathematical Journal 19.1 (1989): 209-219. (Project Euclid)

[2] Bumagin, Inna, and Daniel T. Wise. "Every group is an outer automorphism group of a finitely generated group." Journal of Pure and Applied Algebra 200.1-2 (2005): 137-147. (doi)

[3] Logan, Alan D. "Every group is the outer automorphism group of an HNN-extension of a fixed triangle group." Advances in Mathematics 353 (2019): 116-152. (doi, arXiv)

Solution 1:

Here is a positive answer for finite abelian groups.

For $k > 2$ and any $n > 0$, the simple group $\operatorname{PSp}_{2k}(2^n)$ has outer automorphism group isomorphic to the cyclic group $C_n$. So for any finite cyclic group, there exist infinitely many non-abelian finite simple groups with outer automorphism group $C_n$.

Let $A$ be a finite abelian group, so $A = C_{n_1} \times \cdots \times C_{n_t}$ for some $n_i > 0$ by the fundamental theorem of finitely generated abelian groups.

It follows from the main theorem of [1] that if $G_1$, $\ldots$, $G_t$ are pairwise non-isomorphic nonabelian simple groups, then for $G = G_1 \times \cdots \times G_t$ we have $\operatorname{Out}(G) \cong \operatorname{Out}(G_1) \times \cdots \times \operatorname{Out}(G_t).$ So for example by choosing $G_i = \operatorname{PSp}_{2k_i}(2^{n_i})$ with $2 < k_1 < \cdots < k_t$, we get $\operatorname{Out}(G) \cong A$.

EDIT: To give another example, all symmetric groups are outer automorphism groups. If $H$ is indecomposable and non-abelian with $\operatorname{Hom}(H,Z(H)) = 1$, then by Theorem 3.1 in [2] we have $\operatorname{Aut}(H^n) \cong \operatorname{Aut}(H)^n \rtimes S_n$, where $S_n$ acts on $H^n$ by permuting the direct factors.

Thus $\operatorname{Out}(H^n) \cong S_n$ for example if $H = S_3$, more generally if $H$ is complete (trivial center and all automorphisms inner).

[1] J. N. S. Bidwell, M. J. Curran, and D. J. McCaughan, Automorphisms of direct products of finite groups, Arch. Math. 86, 481 – 489 (2006).

[2] J. N. S. Bidwell, Automorphisms of direct products of finite groups II, Arch. Math. 91, 111–121 (2008).