# Topological semi-direct products of groups

In Kaniuth, Taylor, Induced representations of locally compact groups on pages 9-10 it's claimed that if $$G$$ is a locally compact group with closed subgroups $$N,H$$, with $$N$$ normal in $$G$$, with $$N\cap H=\{e\}$$, and with $$NH=G$$, then $$G$$ is a topological semidirect product of $$N$$ and $$H$$.

We copy the algebraic proof, defining an action $$\alpha_h(n) = hnh^{-1}$$ which will be suitably continuous, allowing us to construct $$N \rtimes_\alpha H$$. The map $$N \rtimes_\alpha H \rightarrow G; (n,h) \mapsto nh$$ is an isomorphism of groups, and clearly continuous.

Why is the inverse of this map continuous?

You would need to show that given nets $$(n_i)\subseteq N, (h_i)\subseteq H$$ with $$n_ih_i\rightarrow e$$, then necessarily $$n_i\rightarrow e, h_i\rightarrow e$$. I don't see how to do this.

(Under some conditions, e.g. that $$N \rtimes_\alpha H$$ is $$\sigma$$-compact, there are open mapping theorems for locally compact groups available, which would show this. For example, see Corollary 1.7 in Hofmann, Morris, Open Mapping Theorem for Topological Groups (pdf).)

let $$K$$ be an infinite compact group, and $$K^\delta$$ be $$K$$ with the discrete topology.
Let $$G$$ be $$K^\delta\times K$$, let $$N$$ be equal to $$K^\delta\times\{0\}$$ and let $$H$$ be the diagonal. Then both $$N,H$$ are discrete, $$G=NH$$, $$N\cap H=\{1\}$$. But the canonical continuous group isomorphism $$N\rtimes H\to G$$ is not a topological isomorphism, since $$G$$ is not discrete.