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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?

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    $\begingroup$ 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. $\endgroup$ – Servaes Nov 17 '13 at 9:55
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    $\begingroup$ 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 $\mathrm{Aut}(G)$. $\endgroup$ – Zhen Lin Nov 17 '13 at 9:57
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    $\begingroup$ 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 $\mathrm{Aut}(H) \to \mathrm{Aut}(NH)$. $\endgroup$ – Martin Brandenburg Nov 17 '13 at 10:44
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Nov 17 '13 at 11:54
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    $\begingroup$ @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 $5$. $\endgroup$ – Servaes Nov 18 '13 at 2:16
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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.

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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 !

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    $\begingroup$ Yah, that's the problem $\endgroup$ – user18537 Jun 17 '15 at 12:51
  • $\begingroup$ Generally, if we can make $|Aut(G_1)|=n, |Aut(G_2)|=m$, it also can deduce a contradiction. $\endgroup$ – user18537 Jun 17 '15 at 12:54

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