# Semidirect product induced by $\varphi:H \rightarrow \textrm{Aut}(N)^{op}$

Suppose $$N,H$$ are groups. Assuming $$\varphi$$ is a homomorphism from $$H$$ to $$\textrm{Aut}(N)$$, we can construct what's called the semi-direct product $$N \rtimes H$$, defining the operation on the cartesian product $$N \times H$$ as $$(n,h) (n',h') = (n \varphi_{h} (n'), hh')$$.

By a direct check we can see that this truly defines a group which is generally denoted by $$N \rtimes H$$, and if we identify $$N$$ with $$\{ (n,0) : n \in N \}$$ and $$H$$ with $$\{ (0,h) : h \in H \}$$, we see that $$N$$ is normal in $$N \rtimes H$$, $$N \cap H =1$$, and $$NH = N \rtimes H$$.

This gives us a way to describe the situation when $$N,H$$ are subgroups of $$G$$, $$N$$ is normal in $$G$$, $$N \cap H = 1$$, and $$NH = G$$ - here we consider the the action of $$H$$ on $$G$$ by conjugation. But thinking about this, I realized that the actual conjugation action is somewhat arbitrarily chosen as either left or right action - conjugation of $$n$$ by $$h$$ can be either $$n \mapsto h n h^{-1}$$ or $$n \mapsto h^{-1} n h$$. If this map is to be a homomorphism from $$H$$ to $$\textrm{Aut}(N)$$ then only the first map works, but I would have expected both of these conjugations to work (in the sense that both can be used to describe some sort of semidirect product of $$N$$ and $$H$$).

If both of these conjugations actually work, this could indicate that a homomorphism from $$H$$ to $$\textrm{Aut}(N)^{op}$$ could also be used to create a semi-direct product of some sort. Here, by $$\textrm{Aut}(N)^{op}$$ I mean the group of automorphisms on $$N$$, where the composition of functions is reversed - i.e. $$\varphi : H \rightarrow \textrm{Aut}(N)^{op}$$ satisfies $$\varphi(ab)=\varphi_{ab}=\varphi_b \circ \varphi_a$$, where $$\circ$$ is the standard composition defined as $$f \circ g (x) = f(g(x))$$.

In fact, if $$\varphi \colon H \rightarrow \textrm{Aut}(N)^{op}$$ is such a homomorphism, then defining the operation on the cartesian product $$N \times H$$ as:

$$(n,h)(n',h') = ( \varphi_{h'}(n) n', hh')$$

seems to work properly - it is a group $$G$$ that contains $$N$$ as a normal subgroup, $$N \cap H =1$$ and $$NH = G$$. Also notice the first coordinate is the same as the opposite group of $$N \rtimes H$$ as constructed by the first method mentioned in this post (first paragraph of this post).

Assuming I haven't made a mistake checking the properties above, why doesn't the semidirect product definition not contain this method of constructing it? Maybe these semi-direct products defined using $$\varphi:H \rightarrow \textrm{Aut}(N)^{op}$$ are all isomorphic to some other semi-direct product constructed by some homomorphism $$\psi:H \rightarrow \textrm{Aut}(N)$$, and so they don't describe anything new.

I would appreciate any input on this issue or reference to a book that considers this issue.

$$\renewcommand{\phi}{\varphi}\DeclareMathOperator{\Aut}{Aut}$$Let $$N, H$$ be groups, and $$\Aut(N)$$ the automorphism group of $$N$$. Let us decide how we can compose on $$\Aut(N)$$, let us say we do it left-to-right, that is, if $$\phi, \psi \in \Aut(N)$$, then in the composition $$\phi \circ \psi$$ it is $$\phi$$ that acts first. In $$\Aut(N)^{op}$$ composition will be right-to-left, then.
Consider a homomorphism $$f : H \to \Aut(N)$$, and construct the semidirect product $$H \ltimes N$$ with multiplication $$(h_{1}, n_{1}) (h_{2}, n_{2}) = (h_{1} h_{2}, n_{1}^{f(h_{2})} n_{2}).$$ (I write automorphisms as exponents.)
With the identifications you have provided, in this group we have $$h^{-1} n h = n^{f(h)}.$$
Now consider the map $$f'$$ given by $$f'(h) = f(h^{-1})$$. It is easily seen that $$f' : H \to \Aut(N)^{op}$$ is a homomorphism. Moreover, it is easily seen that every homomorphism $$H \to \Aut(N)^{op}$$ is of the form $$f'$$, for some homomorphism $$f : H \to \Aut(N)$$.
Now consider the previous semidirect product, and consider in it, with your identifications, $$h n h^{-1} = (h, 1) (n, 1) (h^{-1}, 1) = (h, n) (h^{-1}, 1) = (1, n^{f(h^{-1})}) = n^{f'(h)}.$$ So $$f, f'$$ give the same semidirect product, it is just that you read the action of $$f(h)$$ as a conjugation $$n \mapsto h^{-1} n h$$ (that is, a right action, as it corresponds to left-to-right composition in $$\Aut(G)$$), and that of $$f'(h)$$ as $$n \mapsto h n h^{-1}$$ (that is, a left action, as it corresponds to right-to-left composition in $$\Aut(G)^{op}$$)).