# Taking the automorphism group of a group is not functorial.

Once upon a time I proved that there is no functorial 'association' $$F:\ \mathbf{Grp}\ \longrightarrow\ \mathbf{Grp}:\ G\ \longmapsto\ \operatorname{Aut}(G).$$ A few days ago I casually mentioned this to someone, and was asked for a proof. Unfortunately I could not and still can not recall how I proved it. Here is how much of my proof I do recall:

Suppose such a functor does exist. Choose some group $G$ wisely, and let $f\in\operatorname{Hom}(V_4,G)$ and $g\in\operatorname{Hom}(G,V_4)$ be such that $g\circ f=\operatorname{id}_{V_4}$. Then, because $F$ is a co- or contravariant functor we have $$F(g)\circ F(f)=F(g\circ f)=F(\operatorname{id}_{V_4})=\operatorname{id}_{\operatorname{Aut}(V_4)},$$ or $$F(f)\circ F(g)=F(g\circ f)=F(\operatorname{id}_{V_4})=\operatorname{id}_{\operatorname{Aut}(V_4)},$$ where $\operatorname{Aut}(V_4)\cong S_3$. In particular $\operatorname{Aut}(G)$ contains a subgroup isomorphic to $S_3$. Then something about the order of $\operatorname{Aut}(G)$ leads to a contradiction.

I cannot for the life of me find which goup $G$ would do the trick. Any ideas?

• There is no obvious candidate for what the induced maps should be. The point is to prove that for any choice of induced maps, the association is not functorial. Nov 17, 2013 at 9:55
• It's a nice idea, but seems to require a group $G$ with a rather unusual property. It appears you are looking for a group $G$ with a subgroup $H$ such that the inclusion $H \hookrightarrow G$ has a retraction, while $\mathrm{Aut}(H)$ is a group of greater cardinality than $\mathrm{Aut}(G)$. Nov 17, 2013 at 9:57
• If $NH$ is a semi-direct product, then $H \to NH$ is a split monomorphism, and this property is preserved by any functor. Therefore I would try to find an example of a semi-direct product such that there is no split monomorphism $\mathrm{Aut}(H) \to \mathrm{Aut}(NH)$. Nov 17, 2013 at 10:44
• You could take $NH$ to be a Frobenius group of order 56, with $|N|=8$, $|H|=7$. Then ${\rm Aut}(H)$ is cyclic of order 6, whereas ${\rm Aut}(NH) = NHT$ with $T$ cyclic of order 3. So a generator of ${\rm Aut}(H)$ does not extend to an automorphism of $NH$. ${\rm Aut}(NH)$ does have elements of order 6, but they are not complemented. Nov 17, 2013 at 11:54
• @Derek Holt: Your example would do the trick if the identity on $\operatorname{Aut}(H)$ cannot factor over $\operatorname{Aut}(NH)$. In this case it can, though as far as I know there is no 'nice' way for it to factor. Your general idea is a good one though. I have understood $NH\cong\operatorname{GA}(1,8)$, the group of affine transformations of $\Bbb{F}_8$. The group $\operatorname{GA}(1,32)$ does work; we have $\operatorname{GA}(1,32)=N\rtimes H$ with $|N|=32$, $|H|=31$. Then $\operatorname{Aut}(H)$ is cyclic of order $30$, and $|\operatorname{Aut}(NH)|=NHT$ with $T$ cyclic of order $5$. Nov 18, 2013 at 2:16

Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake.

Let $N:=\Bbb{F}_{11}$ the finite field of $11$ elements and $H:=\Bbb{F}_{11}^{\times}$ its unit group. Let $G:=N\rtimes H$ the semidirect product of $N$ and $H$ given by the multiplication action of $H$ on $N$. Note that $G$ is isomorphic to the group of affine transformations of $\Bbb{F}_{11}$. Then the maps $$f:\ H\ \longrightarrow\ G:\ h\ \longmapsto\ (0,h)\qquad\text{ and }\qquad g:\ G\ \longrightarrow\ H:\ (n,h)\ \longmapsto\ h,$$ are group homomorphisms and satisfy $g\circ f=\operatorname{id}_H$. Now suppose there exists a covariant functor $$F:\ \mathbf{Grp}\ \longrightarrow\ \mathbf{Grp}:\ X\ \longmapsto\ \operatorname{Aut}(X).$$ Then we have group homomorphisms $$F(f):\ \operatorname{Aut}(H)\ \longrightarrow\ \operatorname{Aut}(G)\qquad\text{ and }\qquad F(g):\ \operatorname{Aut}(G)\ \longrightarrow\ \operatorname{Aut}(H),$$ satisfying $$F(g)\circ F(f)=F(g\circ f)=F(\operatorname{id}_H)=\operatorname{id}_{\operatorname{Aut}(H)},$$ so the identity on $\operatorname{Aut}(H)$ factors over $\operatorname{Aut}(G)$, i.e. $\operatorname{Aut}(G)$ has a subgroup isomorphic to $\operatorname{Aut}(H)$.

We have $\operatorname{Aut}(H)\cong\Bbb{Z}/4\Bbb{Z}$ because $H$ is abelian of order $10$. By this question we have $\operatorname{Aut}(G)\cong G$. But $|\operatorname{Aut}(G)|=|G|=11\times10=110$ is not divisible by $|\operatorname{Aut}(H)|=|\Bbb{Z}/4\Bbb{Z}|=4$, a contradiction. This shows that no such covariant functor exists. The contravariant case is entirely analogous.

Maybe, you can consider $G_1=G$ and $G_2=G\oplus H$

Clearly, there are two canonical morphisms, the injection $i:G_1\rightarrow G_2$ and the projection $\pi:G_2\rightarrow G_1$

Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying

1. $Aut(G)$ has a generator $g_1$ of order $n$

2. $Aut(G\oplus H)$ has a generator $g_2$ of order $m$

3. $(n,m)=1$

Then the Lagrange Theorem provides for contradiction

As $\pi\circ i=1_G$, we get

1) covariant case

$\mathrm{ord}(Aut(i)(g_1))|\mathrm{ord}(g_1)=n$, $\mathrm{ord}(Aut(i)(g_1))|m$

Thus, $Aut(i)(g_1)=g_2^0=e_{Aut(G_2)}$, which contradicts to

$Aut(\pi)\circ Aut(i)=1_{Aut(G)}$

2) contravariant case

Change $Aut(i)(g_1)$ to $Aut(\pi)(g_1)$ and it's OK !

• Yah, that's the problem Jun 17, 2015 at 12:51
• Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction. Jun 17, 2015 at 12:54