Semi-Direct Products

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
$((h , g) (h', g'))(h'', g'') = (h \phi(g)(h'), g g') (h'', g'') = (h \phi(g)(h') \phi(g g') (h'') , g g' g'' )$
and
$(h , g) ((h', g')(h'', g'')) = (h, g) (h'' \phi(g')(h''), g' g'') = (h \phi(g) (h'' \phi(g')(h'')), g g' g'')$
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 $G \uparrow \text{Grp}$ :
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?

Semi-Direct Products

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Semi-Direct Products

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