On the objects of the category of groups we define the mapping $G\mapsto \operatorname{Hol}(G)$, the holomorph $G\rtimes \operatorname{Aut}(G)$ of $G$. Can we extend this mapping to a functor on this category? (Via extension to morphisms)



Is $G\mapsto \operatorname{Hol}(G)$ the object component of any functor on the category of groups?
There is no such functor. Recall that a *split epimorphism* is a morphism $f : x \to y$ with a section (right inverse) $g : y \to x$, satisfying $fg = \text{id}\_y$. Split epimorphisms, as their name suggests, are epimorphisms, and moreover they are *absolute* epimorpisms in that they are preserved by any functor whatsoever.


In $\text{Grp}$ every split epimorphism arises as a projection $N \rtimes G \to G$ where $N \rtimes G$ is a semidirect product, so any functor from groups to groups must preserve these. In particular if $\text{Hol}(-)$ were such a functor it would follow that $\text{Hol}(N \rtimes G)$ admits a split epimorphism to $\text{Hol}(G)$.


No such split epimorphism exists in general. Explicitly, take $N = C\_2, G = C\_2^3$ with the trivial action. Then


$$\text{Hol}(N \rtimes G) \cong C\_2^4 \rtimes GL\_4(\mathbb{F}\_2)$$


while


$$\text{Hol}(G) \cong C\_2^3 \rtimes GL\_3(\mathbb{F}\_2).$$


$GL\_n(\mathbb{F}\_2) \cong SL\_n(\mathbb{F}\_2) \cong PSL\_n(\mathbb{F}\_2)$ is simple for $n \ge 3$, so these two groups each have a unique nonabelian simple group in their Jordan-Holder decompositions, namely $GL\_4(\mathbb{F}\_2)$ and $GL\_3(\mathbb{F}\_2)$ respectively. Any non-solvable quotient of $C\_2^4 \rtimes GL\_4(\mathbb{F}\_2)$ must also contain $GL\_4(\mathbb{F}\_2)$ in its Jordan-Holder decomposition, and since $GL\_3(\mathbb{F}\_2)$ and $GL\_4(\mathbb{F}\_2)$ have different orders they are non-isomorphic, so $\text{Hol}(G)$ is not a quotient (or even a subquotient) of $\text{Hol}(N \rtimes G)$.


A similar but simpler argument shows that there is no functor sending a group $G$ to its center $Z(G)$, since for example $\text{sgn} : S\_3 \to C\_2$ is a split epimorphism but there is no epimorphism $Z(S\_3) \to Z(C\_2)$ since the former is trivial and the latter is not. We can also show that there is no functor sending a group $G$ to its automorphism group $\text{Aut}(G)$ using the same counterexample $N = C\_2, G = C\_2^3$ as above, although $N = C\_2, G = C\_2^2$ also works and requires a slightly different argument. I used a similar but more complicated argument on [math.SE recently](https://math.stackexchange.com/questions/3843808/does-the-association-v-mapsto-glv-define-a-functor/3845230#3845230) to show that there is no functor sending a finite-dimensional vector space $V$ to $GL(V)$.


