Posted on 4th February 2019

Motivated by some reading about quantum groups, I want to sketch how a semi-direct product of (topological) groups is the same as having an idempotent group homomorphism.

Firstly, let's remember what the (external) semi-direct product of groups is. I will follow the notation of the book of Kaniuth and Taylor. Let \( N,H \) be (topological) groups, and denote by \( \newcommand{\aut}{\operatorname{Aut}}\aut(N) \) the collection of continuous group automorphisms of \( N \). Suppose we have a group homomorphism \( \alpha:H\rightarrow\aut(N) \), written as \( h\mapsto \alpha_h \), which is continuous in the sense that \( N\times H\rightarrow N; (n,h)\mapsto \alpha_h(n) \) is continuous.

Let \( N \rtimes H \), the semi-direct product, be the topological space \( N\times H \) equipped with the group product \[ (n,h)(m,g) = (n \alpha_h(m), hg). \] It is easy to check that this is associate, that \( (e,e) \) is the unit, and that \( (n,h)^{-1} = (\alpha_{h^{-1}}(n^{-1}), h^{-1}) \). Similarly, it is easy to check that the product and inverse are continuous, so that \( N \rtimes H \) becomes a topological group.

Let \( \tilde N = \{(n,e) : n\in N\} \) and \( \tilde H = \{(e,h):h\in H\} \). These can be checked to be closed subgroups of \( N \rtimes H \) such that \( \tilde N \cap \tilde H = \{e\} \) and \( \tilde N\tilde H = N \rtimes H \). For \( (m,e)\in \tilde N \) and \( (n,h)\in N \rtimes H \) we have that \[ (n,h)(m,e)(n,h)^{-1} = (n\alpha_h(m), h)(\alpha_{h^{-1}}(n^{-1}), h^{-1}) = (n\alpha_h(m)n^{-1}, e), \] which shows that \( \tilde N \) is normal in \( N \rtimes H \), and that \( (\alpha_h(m),e) = (e,h)(m,e)(e,h)^{-1} \), so that conjugation by \( \tilde H \) in \( N \rtimes H \) implements \( \alpha \).

Conversely, suppose that \( G \) is a topological group, with closed subgroups \( N,H \), with \( N \) normal, \( N\cap H=\{e\} \) and \( G = NH \). Define \( \alpha:H\rightarrow\aut(N) \) by \( \alpha_h(n) = hnh^{-1} \) for \( h\in H, n\in N \). As \( N \) is normal, it follows that \( \alpha \) makes sense, and is a (continuous) group homomorphism. Consequently, we can define \( N \rtimes H \).

Define \( \theta:N \rtimes H\rightarrow G \) by \( \theta(n,h)=nh \). Then \[ \theta(n,h)\theta(m,g) = nhmg = nhmh^{-1}hg = \theta(n\alpha_h(m), hg), \] showing that \( \theta \) is a group homomorphism. By assumption, \( \theta \) is onto, and if \( nh=e \) then \( n = h^{-1} \in N\cap H \) so \( n=h=e \), so \( \theta \) is a bijection. Thus \( G \) is isomorphic to \( N \rtimes H \).

We now show the (I think less common) link with idempotent group homomorphisms. With \( G,N,H \) as before, for \( g\in G \) we can write \( g=nh \). This decomposition is unique, either by the above map \( \theta \), or by observing that if \( nh=mg \) then \( m^{-1}n = gh^{-1} \in N\cap H \) so \( m^{-1}n=e \) and \( gh^{-1}=e \) so \( n=m, g=h \). Define \( \phi : N \backslash G \rightarrow H \) by \( \phi(Ng) = h \) when \( g=nh \) for some \( n\in N \). Notice that \( g=nh \) for some \( n\in N \) is equivalent to \( Ng = Nh \). Then \( \phi(Nh) = h \) for \( h\in H \) so \( \phi \) is onto, and if \( \phi(gN) = e \) then \( g\in N \) so \( gN=N \). As \( N \) is normal, of course \( Ng_1 Ng_2 = N g_1 g_1^{-1}Ng_1 g_2 = N g_1g_2 \). Let \( \phi(Ng_1)=h_1 \) and \( \phi(Ng_2)=h_2 \), so that \( Ng_1=Nh_1 \) and \( Ng_2=Nh_2 \), so \( Ng_1g_2 = Ng_1 Ng_2 = Nh_1 Nh_2 = Nh_1h_2 \). Thus \( \phi \) is a group isomorphism. To argue that \( \phi \) is continuous seems a little tricky, but it follows rather easily if you view \( G \) as being isomorphic to \( N \rtimes H \).

Now let \( p \) be the composition \[ G \twoheadrightarrow N \backslash G \cong H \hookrightarrow G \] which is a (continuous) group homomorphism. Thus \( p(g)=h\in H \) if and only if \( h = \phi(Ng) \), that is, \( Nh=Ng \) or equivalently, \( gh^{-1} \in N \). Thus \( p(h)=h \) for any \( h\in H \), so the image of \( p \) is \( H \), and the kernel of \( p \) is \( N \). In particular, \( p\circ p=p \).

Conversely, let \( G \) be a topological group, and let \( p \) be a continuous group homomorphism from \( G \) to \( G \), with \( p\circ p =p \). Let \( N=\ker p \), a closed normal subgroup of \( G \), and let \( H \) be the image of \( p \), a subgroup. As \( p\circ p=p \) and \( p \) is continuous, it follows that \( H \) is closed. If \( g\in N\cap H \) then \( p(g)=e \) and \( p(g)=g \), so \( N\cap H=\{e\} \). For \( g\in G \), let \( n = g p(g)^{-1} = gp(g^{-1}) \) so \( p(n) = p(g) p(g^{-1}) = p(g)p(g)^{-1} = e \). Thus \( g = n p(g) \in NH \).

Thus the semi-direct product construction is exactly captured by the notion of a group with an idempotent group homomorphism.