Taking the automorphism group of a group is not functorial.

Once upon a time I proved that there is no functorial ‘association’
F: Grp  Grp: G  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 fHom(V4,G) and gHom(G,V4) be such that gf=idV4. Then, because F is a co- or contravariant functor we have
F(g)F(f)=F(gf)=F(idV4)=idAut(V4),
or
F(f)F(g)=F(gf)=F(idV4)=idAut(V4),
where Aut(V4)S3. In particular Aut(G) contains a subgroup isomorphic to S3. Then something about the order of Aut(G) leads to a contradiction.

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

Answer

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.

Attribution
Source : Link , Question Author : This site has become a dump. , Answer Author : This site has become a dump.

Leave a Comment