Let’s talk about semi-direct products! We’ll talk about the categorical significance of semi-direct products and some simple examples. But first, just what is a semi-direct product?

Take a group G with a structure-respecting group action of G on a group H. In other words, we have a homomorphism \phi : G \rightarrow \text{Aut}(H). Contrast this with a group action of G on H which does not respect the group structure of H, or in other terms a homomorphism G \rightarrow \text{Bij}(H). From \phi we can form a group H \rtimes_{\phi} G. We write H \rtimes G when the context is clear. As a set this is H \times G, but we put an operation on it which is distinct from the usual group product. Instead, we define (h, g) (h', g') = (h \phi(g)(h'), g g'). This group is called the semi-direct product of G and H. Note that it is dependent on the homomorphism \phi. If G acts on H trivially, then we get the ordinary direct product G \times H, since (h, g)(h', g') = (h \phi(g)(h') , g g') = (h h', g g'). But in general, the semi-direct product is ‘twisted’ in a way, and certainly not isomorphic to G \times H.

Let’s check that this is a group. To see the operation is associative, take g, g', g'' \in G and h, h', h'' \in H. Then
((h , g) (h', g'))(h'', g'') = (h \phi(g)(h'), g g') (h'', g'') = (h \phi(g)(h') \phi(g g') (h'') , g g' g'' )
(h , g) ((h', g')(h'', g'')) = (h, g) (h'' \phi(g')(h''), g' g'') =  (h \phi(g) (h'' \phi(g')(h'')), g g' g'')
And the RHS of the two equations is seen to be equal since \phi is a homomorphism and \phi(g) is a homomorphism. (1, 1) is an identity for H \rtimes G: (1, 1) (h, g) = (\phi(1)(h), g) = (h, g) and (h, g)(1, 1) = (h, g). And lastly we need an inverse for (h, g). It will have to be of the form (h, g)(x, g^{-1}) for some x \in H. We need h \phi(g)(x) = e, so that we can take x = \phi(g^{-1})(h^{-1}) and that will do. But we should check that (x, g^{-1})(h, g) = (1, 1): (x, g^{-1})(h, g) = (x \phi(g^{-1})(h), 1) = (\phi(g^{-1})(h^{-1}) \phi(g^{-1})(h), 1) = (1, 1). So H \rtimes G is a group.

Let’s look at a few examples. Take an abelian group A. \mathbb{Z} / 2 \mathbb{Z} acts on A by \overline{1} \cdot a = a^{-1} and \overline{0} \cdot a = a. We need to require that A be abelian, so that \overline{1} \cdot (ab) =(ab)^{-1} = b^{-1} a^{-1} = a^{-1} b^{-1} = (\overline{1} \cdot a) (\overline{1} \cdot b). Under this action, we get a group A \rtimes \mathbb{Z} / 2 \mathbb{Z}. We can describe it with generators and relations as (A * \mathbb{Z} / 2 \mathbb{Z})  / K where K is the subgroup generated by the relations \{ at = ta^{-1} : a \in A \}. When A = \mathbb{Z} /n \mathbb{Z}, we get D_{n}.

Another interesting example: \mathbb{R}^n and O_n. For x \in \mathbb{R}^n and M \in O_n, put M \cdot x = Mx. \mathbb{R}^n \rtimes O_n \cong E(n), the Euclidian group.

It is not hard to see that H \rtimes G \cong (H * G)/ K where K is the smallest normal subgroup containing the relations g h  = \phi(g) (h) g. Take the homomorphism H * G \rightarrow H \rtimes G. It’s surjective, and it can be seen to have kernel K. This gives us a way of expressing the semi-direct product as a colimit, but it’s not the nicest categorical expression at hand. To see the nicer property, let G \uparrow \text{Grp} be the category of groups under G. It’s objects are homomorphisms \phi : G \rightarrow H, and its morphisms \phi \rightarrow \psi are morphisms \alpha : \text{Cod} (\phi) \rightarrow \text{Cod}(\psi) such that \alpha \circ \phi = \psi. We also have the category of G-groups, G \text{-Grp}. Its objects are groups H with morphisms G \rightarrow \text{Aut}(H), and its morphisms are G-equivariant homomorphisms.

We can think of the objects in G \uparrow \text{Grp} as special types of G-groups. Every morphism \phi: G \rightarrow H induces a G-group, which consists of H with the G-action g \cdot h = \phi(g) h \phi(g)^{-1}. If we replace an object \phi : G \rightarrow H under G with its corresponding G  \text{-Grp} \psi : G \rightarrow \text{Aut}(H), then we lose information about the homomorphism, and it can’t be recovered. But there is a “best approximation”, namely G \rightarrow H \rtimes_{\psi} G!

In categorical terms, we have a forgetful functor F : G \uparrow \text{Grp} \rightarrow G \text{-Grp} whose left adjoint - \rtimes G : G  \text{-Grp} \rightarrow G \uparrow \text{Grp} sends objects \phi : G \rightarrow \text{Aut}(H) to objects H \rtimes_{\phi} G. A morphism H \rightarrow H' under G is sent to that same morphism in G  \text{-Grp}, and a G-equivariant homomorphism \alpha : H \rightarrow H' is sent to the morphism \alpha : H \rtimes G \rightarrow H' \rtimes G where (h, g) \mapsto (\alpha(h), g).

To see that this is actually an adjoint relationship, take a G-group H and set a map \eta_H : H \rightarrow H \rtimes G where h \mapsto (h, 1). We’ll check the unit definition of an adjunction: for each object G \stackrel{\phi}{\rightarrow} K in G \uparrow \text{Grp} and each morphism \alpha : H \rightarrow F(K), there is a unique morphism \beta: H \rtimes G \rightarrow K such that F(\beta) \circ \eta_H = \alpha.
Screen Shot 2018-02-03 at 8.50.13 PM
Take an object G \stackrel{\phi}{\rightarrow} K and a G-equivariant homomorphism \alpha : H \rightarrow F(K). We must set \beta : H \rtimes G \rightarrow K where \beta(h, 1) = \alpha(h) and \beta(1, g) = \phi(g). This gives \beta(h, g) = \alpha(h, 1)\phi(1, g). Taking h, h' \in H and g, g' \in G. Then
\alpha(h \phi(g)(h')) \phi(gg') = \alpha( h) \alpha(\phi(g)(h')) \phi(g) \phi(g') = \alpha (h) \phi(g) \alpha(h') \phi(g)^{-1} \phi(g) \phi(g') = \alpha(h) \phi(g) \alpha(h') \alpha(h') \phi(g')
As desired.

There is a nice characterization of which objects G \rightarrow H in G \uparrow H are isomorphic in G \uparrow \text{Grp} to G \rightarrow K \rtimes G for some G-group K. Recall that retract of a morphism G \rightarrow H is a morphism H \rightarrow G such that G \rightarrow H \rightarrow G is the identity. A morphism G \rightarrow H is isomorphic in G \uparrow \text{Grp} to G \rightarrow K \rtimes G for some G-group K if and only if it has a retract. For a G-group K, it is clear that the composition G \stackrel{\phi}{ \rightarrow} K \rtimes G \stackrel{\psi}{\rightarrow} G, where \phi(g)= (1, g), and \psi(1, g) = g, is the identity. Conversely, suppose we have a morphism G \rightarrow H with retract \phi : H \rightarrow G. Let K = \ker(\phi). Since K is normal, the G-group on H where h \mapsto \phi(g) h \phi(g)^{-1} induces a G-group on K. The following diagram commutes, where \iota : K \rightarrow F(H) is the inclusion map: Screen Shot 2018-02-04 at 1.54.42 PM
Note that \iota is G-equivariant by the construction of the action of G on K. Applying the adjoint correspondence, we get a commutative diagram in G \uparrow \text{Grp}: Screen Shot 2018-02-04 at 1.54.37 PM
Examination of the rows reveals that they are in fact exact sequences. That \alpha in the above diagram is an isomorphism follows from the 5-lemma. N.B. the 5-lemma is usually applied in the context of abelian groups or R-modules – or an abelian category, using the Mitchell’s embedding theorem. One can check, however, that the usual diagram chase works here, however. Thus we have an isomorphism K \rtimes G \rightarrow H in G \uparrow \text{Grp}.

Exercise: This observation also allows for a nice internal characterization of a semi-direct product. Suppose that G and N are subgroups of a group H such that (i) NG = H, (ii) N \cap G = \{ e \}, and (iii) N is normal. Show that H / N \cong G, and that there is a section H / N \rightarrow H. What about the converse?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s