# Order of automorphisms of non-abelian group

Problem: prove that $$\bigl| \operatorname{Aut} \, (G) \bigr| \ge 8$$ if $$G$$ is non-abelian and not isomorphic with $$S_3$$.

The beginning of the solution I have the following: given that a non-abelian group has a non-cyclic factor at the center, and therefore there cannot be a simple order, it is sufficient to prove the statement in the case when the factor at the center is isomorphic to $$V_4$$ or $$S_3$$. In the first case, it all comes down to a situation where the group is a 2-group. In the second case, using Hall's theorem on Hall subgroups, everything boils down to a situation where the group has order $$2^a 3^b$$. How to investigate these groups is not very clear.

• What have you tried? Apr 25, 2022 at 21:15
• What exactly is the "it" that "doesn't work"? Apr 25, 2022 at 21:19
• Given that a non-abelian group has a non-cyclic factor at the center, and therefore there cannot be a simple order, it is sufficient to prove the statement in the case when the factor at the center is isomorphic to $V_4$ or $S_3$. In the first case, it all comes down to a situation where the group is a 2-group. What to do next is unclear. Apr 25, 2022 at 21:20
• Please edit the question to include your attempts. Apr 25, 2022 at 21:30
• I think this should be reopened now that the OP has included some attempts to solve the problem. It seems moderately difficult. Apr 26, 2022 at 8:08

As the poster observed, we just need to solve the problem for (finite) groups $$G$$ with $$G/Z(G) \cong S_3$$ or $$V = C_2 \times C_2$$.

Recall that a central automorphism $$\alpha$$ of a group $$G$$ is one that induces the identity on both $$Z(G)$$ and $$G/Z(G)$$. The central automorphisms form a (normal) subgroup of $${\rm Aut}(G)$$ and, for all $$g \in G$$, we have $$\alpha(g) = g \phi(g)$$, where $$\phi:G \to Z(G)$$ is a homomorphism with $$Z(G) \le \ker \phi$$. Conversely, any such $$\phi$$ determines a central automorphism of $$G$$, so the group of central automorphisms is isomorphic to $${\rm Hom}(G/Z(G),Z(G))$$.

$$\textit{Case 1}$$. $$G/Z(G) \cong S_3$$. If $$|Z(G)|$$ is even, then $${\rm Hom}(G/Z(G),Z(G))$$ is nontrivial, so there are nontrivial central automorphisms. The inner automorphisms of $$G$$ are not central, so we have $$|{\rm Aut}(G)| \ge 12$$ and we are done.

So $$|Z(G)|$$ is odd, and $$G$$ has a normal abelian subgroup $$N$$ of odd order with $$|G/N| = 2$$ and $$|N/Z(G)| = 3$$. If $$N$$ is not a $$3$$-group and $$P \in {\rm Syl}_p(N)$$ with $$p \ne 3$$, then $$P$$ is a direct factor of $$G$$ and there are non-inner automorphisms of $$G$$ acting on $$P$$ and again we are done.

So $$N$$ is a $$3$$-group. Let $$x,y$$ be inverse images in $$G$$ of elements of generators of $$G/Z(G)$$, where we can choose $$x$$ to have order $$2$$ and $$y \in N$$. Then there are only three distinct commutators in $$G$$, the identity, $$[x,y]$$ and $$[x,y^2]$$, so $$|[G,G]| = 3$$. Since $$[x,y] \in N \setminus Z(G)$$, we have $$[G,G] \cap Z(G) = 1$$ and hence $$N = [G,G] \times Z(G)$$, and $$G = \langle x,y \rangle \times Z(G)$$. Again there are non-inner automorphisms of $$G$$ acting on $$Z(G)$$, and we are done.

$$\textit{Case 2.}$$ $$G/Z(G) \cong C_2 \times C_2$$. Let $$x,y$$ be inverse images in $$G$$ of generators of $$G/Z(G)$$. The only distinct commutators in $$G$$ are $$1$$ and $$[x,y]$$, so $$|[G,G]|=2$$, and $$[G,G] \le Z(G)$$. Hence $$G$$ is nilpotent, and we reduce as in Case 1 to the case when $$G$$ is a $$2$$-group.

If $$Z(G)$$ is non cyclic, then $$|{\rm Hom}(G/Z(G),Z(G))| \ge 16$$ and we are done, so $$Z(G) = \langle z \rangle$$ is cyclic.

If $$z$$ has order $$2$$ so $$|G|=8$$, then $$G \cong D_8$$ or $$Q_8$$, and $$|{\rm Aut}(G)| = 8$$ or $$24$$, and again we are done.

So $$z$$ has order $$2^k$$ with $$k>1$$. Note that $$x^2$$ and $$y^2$$ are powers of $$z$$. If either of them are even powers of $$z$$, then we can replace them by $$xz^i$$ and/or $$yz^j$$ for suitable $$i,j$$ to get elements of order $$2$$.

If we can do this for both $$x$$ and $$y$$, then $$G$$ is a central product of $$D_8$$ with $$\langle z \rangle$$, and there is a non-inner automorphism fixing $$x$$ and $$y$$ and mapping $$z$$ to $$z^{-1}$$, and we are done.

Otherwise, we have for example $$x^2=z$$. If $$y^2$$ is also an odd power of $$z$$, then $$(xy)^2$$ is an even power, so we can assume that $$y$$ has order $$2$$, and now $$G$$ is the group with the presentation $$\langle x,y \mid x^{2^{k+1}} = y^2=1, y^{-1}xy = y^{2^k+1} \rangle,$$ and $$G$$ has a non-inner automorphism with $$x\mapsto x^{-1}$$ and $$y \mapsto y$$.

• This is ingenious! Apr 29, 2022 at 22:18