Universal property of $N\rtimes K$

Question

Given groups $N$ and $K$, if $K$ acts on $N$ by \begin{equation} K\xrightarrow{\theta}\operatorname{Aut_{Grp}}(N), \end{equation} then we can define a group $N\rtimes_{\theta}K$ whose elements are like in $N\times K$ but the multiplication is defined by \begin{equation} (n,k)(n',k')=(n\theta_{k}(n'),kk') \end{equation} where $\theta_k\in \operatorname{Aut_{Grp}}(N)$ is the image of $k$ under $\theta$.

This semidirect product is very useful in studying structures of finite groups, especially it solves the extension problem \begin{equation} 1\longrightarrow N\longrightarrow N\rtimes_{\theta} K\longrightarrow K\longrightarrow 1. \end{equation} But I am wondering whether this can be defined using universal properties? In the abelian case $N\rtimes K$ is just $N\times K$ so we have the universal property of products, but what about the nonabelian case?

Thanks!

Answer 1

For the sake of the MSE search engine, I'm summarizing the discussion in the associated MO post.

First, note that each group can be considered to be a (relatively simple) category. If we assume that our semidirect factors $N$ and $K$ are encoded this way, then one can encode $\theta$ as a functor $K\to\textbf{Cat}$ sending $\star\mapsto N$. Post-encoding, one obtains a nice description of $N\rtimes K$ as a Grothendieck construction/Kan extension $\int^K{N}$. If you want fancy names, then this describes $N\rtimes K$ as a lax 2-colimit.

Unfortunately, you probably don't just want fancy names: you probably want a description of semidirect products that is element-free. The encoding process to make a lax 2-colimit construction elevates each element of $N$ and $K$ to a category morphism, which defeats the point. So can we give an element-free description?

Yes! Consider the morphism category $\textbf{Mor}(\textbf{Grp})$ which consists of group homomorphisms $A\to B$. An element $F\in\textbf{Mor}(\textbf{Grp})$ defines a map $A\to\text{Aut}(B)$; elements of $A$ act on $B$ by conjugating by the image under $F$. There is a forgetful functor dropping the specific map $A\to B$ in favor of the action $A\to\text{Aut}(B)$; the mapping taking $K\to\text{Aut}(N)$ to the inclusion $K\to N\rtimes K$ is the left-adjoint of that forgetful functor.

Both of these options are discussed on the $n$-CatLab.

Lastly, one might object that both of these constructions build the semidirect product as a colimits. One usually encounters semidirect products are a generalization of direct products, which are limits, so one might hope for a limit construction. But this is something of a coincidence: the direct product is a quotient of the coproduct (free product) via $$N\times H=N*H/\langle\!\langle nhn^{-1}h^{-1}\rangle\!\rangle_{n\in N, h\in H}$$ where $\langle\!\langle\cdot\rangle\!\rangle$ denotes the normal closure generated by those elements. A semidirect product is a twist of this quotient via $$N\times H=N*H/\langle\!\langle nhn^{-1}\theta_h(n)^{-1}\rangle\!\rangle_{n\in N, h\in H}$$ Indeed, within the category of groups (or, really, any subcategory of pointed sets), a direct product has both projection and injection maps to and from (respectively) each direct factor. The projections are what makes it a direct product…and the condition that semidirect products choose to weaken.

Answer 2

Since this got bumped already today, I'll add another answer based on Martin Brandenburg's answer in the linked question. The point of this answer is to give a short and to the point answer that's (ideally) a little more transparent to beginners than the other existing answer.

If $\phi:K\to \newcommand\Aut{\operatorname{Aut}}\Aut(N)$, then $N\rtimes_\phi K$ is universal among groups $G$ with pairs of maps $i : N\to G$ and $j : K\to G$ such that for all $n\in N$, $k\in K$, $$ j(k)i(n)j(k)^{-1} = i(\phi_k(n)), $$ where $\phi_k=\phi(k)$, In the sense that for any such group $G$ there is a unique map $(i,j) : N\rtimes_\phi K \to G$ such that $(i,j)|_N = i$, and $(i,j)|_K = j$.

In words, $N\rtimes_\phi K$ is universal (initial) for groups $G$ with maps from $N$ and $K$ where the image of $K$ is in the normalizer of the image of $N$, the kernel of $N\to G$ is fixed (setwise) by the action of $K$, and the conjugation action of $K$ on the image of $N$ is given by $\phi$.

As discussed in the other answer, despite products being a special case of semidirect products, we only have injection maps and not projections for semidirect products, so they are a kind of colimit.

For other perspectives and more details read the other answer by Jacob Manaker or the MO answer by Martin Brandenburg.