Products, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits.

As was pointed in: http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/

Bourbaki (General Topology, Prop. 27) gives a universal property:

Let $f \colon N \to G$, $g \colon H \to G$ be two homomorphisms into a group $G$, such that $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ for all $n \in N$, $h \in H$. Then there is a unique homomorphism $k \colon N \rtimes H \to G$ extending $f$ and $g$ in the usual sense.

However, I remain unsatisfied. The condition $f(\phi_h(n)) = g(h)f(n)g(h^{-1})$ is a condition on elements of groups, rather than a condition that says that some diagram is commutative.

So the question remains: are semi-direct products in the category of groups categorical limits?

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    $\begingroup$ It's a certain colimit. Do you know the Grothendieck construction for fibrations? $\endgroup$ May 5, 2012 at 17:43
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    $\begingroup$ Colimit? Are you sure about what you're saying? After all products are a particular case of semi-direct products, and they are limits not colimits. I don't know anything about Grothendieck's construction for fibrations... $\endgroup$ May 5, 2012 at 17:46
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    $\begingroup$ Sure, colimit, because you describe maps on the semi-direct product. Actually this universal property of the semi-direct products in the special case of products does not give you the usual universal property of a categorical product: It gives you that group morphisms $N \times H \to G$ are given by pairs of group morphisms $N \to G$, $H \to G$ which commute pointwise. In other words, $N \times H = N * H / \langle\langle nhn^{-1} h^{-1} \rangle\rangle$, and here you already see the colimit. In a general semi-direct product, this commutator is twisted. $\endgroup$ May 5, 2012 at 17:52
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    $\begingroup$ Semi-direct products involve an action $\phi: H \to Aut(N)$ in addition to the two factor groups $H,N$. So the first step would be to figure out how to describe this action in purely category-theoretic terms, without referencing individual elements. I don't know how to do this, but suspect that if one can achieve this, then describing the semi-direct product category-theoretically should be straightforward. $\endgroup$
    – Terry Tao
    May 5, 2012 at 18:09
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    $\begingroup$ Semidirect products of profinite groups are special because a compactness argument lets you prove one can obtain the action as an inverse limit of actions of finite quotients. $\endgroup$ May 5, 2012 at 21:47

2 Answers 2


This is a partial answer, summing up some of my comments.

The semi-direct product is not a limit, but rather it is a colimit. The reason is that the universal property cited above describes maps on the semi-direct product. In the special case that $\phi$ is the trivial action, the semi-direct product becomes the direct product $N \times H$ and the universal property is not just the usual universal property as a product, but rather as a representing object of the pairs of morphisms on $N,H$ which commute pointwise. In a general semi-direct product, this commutation is twisted by an action of $H$ on $N$.

So basically the idea is that we have the coproduct $N * H$ of the two groups (which is usually called the free product, which is quite unfortunate), and we impose the relation $h n h^{-1} = \phi_h(n)$. The universal property of $N \rtimes H$ is equivalent to the isomorphism

$$N \rtimes H = (N * H) / \{h n h^{-1}= \phi_h(n)\}_{h \in H, n \in N},$$

which exhibits $N \rtimes H$ as a special colimit of some diagram associated to $N,H,\phi$. However, this still uses elements in the relations. I think we cannot get rid of these elements, unless we use $2$-colimits. See below. Actually this isomorphism is used very often in group theory in order to recoqnize groups given by some presentation as a semi-direct product. For example, the dihedral group $D_n = \langle r,s : r^n = s^2 = 1, srs=r^{-1} \rangle$ is $\mathbb{Z}/n \rtimes \mathbb{Z}/2$.

On the other hand, there is a purely category-theoretic construction which is due to Grothendieck: Let $I$ be a small category and $F : I \to \mathsf{Cat}$ be a diagram of small categories. The Grothendieck construction $\int^I F$ is the category of pairs $(i,x)$, where $i$ is an object of $I$ and $x$ is an object of $F(i)$. A morphism $(i,x) \to (j,y)$ is a pair $(a,f)$, consisting of a morphism $f : i \to j$ and a morphism $a : F(f)(x) \to y$ in $F(j)$. The composition is defined by the rule

$(a_2,f_2) \circ (a_1,f_1) = (a_2 \circ F(f_2)(a_1),f_2 \circ f_1)$.

Now if $H$ is a monoid, considered as a category with just one object $*$, and $F : H \to \mathsf{Cat}$ is a diagram such that $F(*)=N$ is just a monoid, then $F$ corresponds to a homomorphism of monoids $H \to \mathrm{End}(N)$ and the Grothendieck construction $\int^H N$ has just one object, thus corresponds to a monoid, namely what is usually called the semi-direct product $N \rtimes H$. This is shown by the multiplication rule above.

Back to the general case of a diagram $F : I \to \mathsf{Cat}$, the Grothendieck construction $\int^I F$ is the lax 2-colimit of $F$. I don't know the original reference right now, but a very comprehensive account on that is the Appendix A in "The stack of microlocal sheaves" by I. Waschkies. The choice of the morphism $a : F(f)(x) \to y$ in the definition above is precisely the reason for the "2" here. If it was the identity, we would get the usual colimit.

Thus, the semi-direct product $N \rtimes H$ is the lax $2$-colimit of the diagram $N : H \to \mathrm{Cat}$.  

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    $\begingroup$ In your answere $H$ have to be a group (instead change $Aut(N)$ by $End(N)$). If $H$ and $N$ are internal groups, and we can internalizing $Aut(N)$ with a natural map $Aut(N)\times N\to N$ (this is possible in enought good enriched $V$-categories), then the semidirected composition has a internal formulation, furthermore I think that a action $\phi: L\to Aut(L)$ could be a pseudo-functor (consider groups as 2-cetegories, with one object on one (identity) morphism), then I guess that the semidirect-products is some kind of a (co)lax-colimit ...(sorry for my bad English) $\endgroup$ May 5, 2012 at 20:20
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    $\begingroup$ If you want, a traditional source on limits (and variations of these) on 2-categories is J. W. Gray "adjointness for 2-categories" (LNM 391). $\endgroup$ May 6, 2012 at 10:35

There is (another ?) description of the crossed product in categorical terms.

Let ${\rm Mor}(Gp)$ be the category whose objects are homomorphisms of groups and morphisms are commutative diagrams. Let $C$ be the category of "groups acting on groups" whose objects are pairs of groups $(H,G)$ together with a homomorphism $H \to {\rm Aut}(G)$. Morphisms in this category are equivariant homomorphisms.

Now, there is a natural forgetful functor $T \colon {\rm Mor}(Gp) \to C$ which sends $H \to G$ to the pair $(H,G)$ with the homomorphism $H \to {\rm Aut}(G)$ given by conjugation. Now, almost by definition, the crossed product is the left-adjoint of this forgetful functor. Indeed, the left adjoint is easily seen to map $(H,G)$ with $H \to {\rm Aut}(G)$ to the inclusion $H \to G \rtimes H$.

Being a left-adjoint, the "crossed product" maps colimits to colimits.

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    $\begingroup$ This is a very concise categorical description! $\endgroup$ May 3, 2013 at 15:12
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    $\begingroup$ Just checking: are the morphisms in $C$ just homomorphisms $(H, G) \to (H, G')$, or are we permitted to vary the group which is acting as well? That is, can we have a pair of group homomorphism $\phi: H \to H'$ and $\psi: G \to G'$ such that $\psi(h.g) = \phi(h).\psi(g)$ as a morphism in $C$? $\endgroup$
    – ಠ_ಠ
    Mar 10, 2016 at 11:27
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    $\begingroup$ Yes, exactly, $H$ need not be fixed. $\endgroup$ Mar 10, 2016 at 14:34
  • $\begingroup$ No, but the map from G to G' can be equivariant with respect to phi. $\endgroup$ Sep 14, 2022 at 11:12
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    $\begingroup$ Yes. It is phi x psi on the top arrow. $\endgroup$ Sep 14, 2022 at 13:10

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