In the case of outer (synthetic) semi-direct products, we take any two groups $N$ and $H$ and a group homomorphism $\varphi: H \to \mathrm{Aut}(N)$ and effectively "synthesize" a new group named $(N \rtimes _\varphi H)_\text{synthetic}$, with the underlying set as the cartesian product $N\times H$ and with a restriction imposed in form of a new group operation, by the homomorphism $\varphi$, i.e.,
$$\bullet: (N \rtimes_\varphi H)_\text{synthetic} \times (N\rtimes_\varphi H)_\text{synthetic}\to (N\rtimes_\varphi H)_\text{synthetic}$$ and just like in the descriptive case
$${(n_{1},h_{1})\bullet (n_{2},h_{2})=(n_{1}\varphi (h_{1})(n_{2}),\,h_{1}h_{2})=(n_{1}\varphi _{h_{1}}(n_{2}),\,h_{1}h_{2})}.$$
Say the identity element in the group is $(1_N, 1_H)$ and the inverse of an element $(n, h)$ is $(\varphi_{h^{-1}}(n^{-1}), h^{-1})$. Now the pairs $(n, 1_H)$ form a normal (*) subgroup $\mathcal{N} \cong N$ and the pairs $(1_N, h)$ form a subgroup $\mathcal H \cong H$. The descriptive semidirect product of these two subgroups $\mathcal{N} \rtimes_\varphi \mathcal{H}$ is in fact the whole artificially constructed group $(N \rtimes_\varphi H)_\text{synthetic}$, in the same sense of inner semi-direct products.
Question:
Unlike in inner semi-direct product definitions, I never see the condition $N \cap H = \{1\}$ in the definition of outer semi-direct products. Why is that?
Is $\mathcal{N} \cap \mathcal{H} = \{1_N, 1_H\}$ by construction? I'm not sure how to prove it.
Best Answer
Yes. $\mathcal{N}$ is equal to $N\times\{1_H\}$, and $\mathcal{H}$ is equal to $\{1_N\}\times H$. The intersection of these two sets is $\{e\}$, where $e=(1_N,1_H)$ is the identity of $N\rtimes_\varphi H$.
The reason we can't stipulate $N\cap H=\{1\}$ in advance is that "$1$" doesn't refer to anything concrete, at least not before you make further identifications. For example $N$ could be a set of integers and $H$ could be a set of automorphisms of some Riemannian manifold. How would you intersect them to get something meaningful? You have to make an identification first. This identification is almost exactly what $N\rtimes_\varphi H$ accomplishes.