Semidirect product: general automorphism always results in a conjugation When $G$ is a group, $N$ is a normal subgroup of $G$ and $H$ is another subgroup of $G$ where $ N \cap H = \{1\} $, the normality of $N$ suggests that we can write, for $n_1, n_2 \in N$ and $h_1, h_2 \in H$,
$$ n_1 h_1 n_2 h_2 = n_1 h_1 n_2 h_1^{-1} h_1 h_2 $$
and so motivates the definition of an 'external' semidirect product using
$$ (n_1,h_1) (n_2,h_2) = (n_1 h_1 n_2 h_1^{-1}, h_1 h_2). $$
However, in general there is no reason to suppose $\textit{a priori}$ that $N$ and $H$ are subgroups of a larger group $G$, so that in general we say that to form the external product we need some groups $N$, $H$, and some homomorphism $\phi \colon H \to \textrm{Aut}(N)$ and define
$$ (n_1,h_1) (n_2,h_2) = (n_1 \phi(h_1)(n_2), h_1 h_2). $$
I would expect that this would give something more general than the intuitive external product given above, since now we are using the result of a general automorphism $ \phi(h_1)(n_2) $ rather than the specific conjugation $ h_1 n_2 h_1^{-1} $. But it turns out that for any $\phi$ you come up with, this defines conjugation in the group $ N \rtimes H$.
I am having difficulty seeing why this is intuitively. Is there any insight anyone can give? Why must the general automorphism in the external construction always correspond to an inner automorphism conjugation in the internal construction?
 A: Use the same idea. Observe
$$\begin{array}{ll} (n_1,h_1)(n_2,h_2) & =(n_1,e_H)(e_N,h_1)(n_2,e_H)(e_N,h_2) \\ & = (n_1,e_H)\color{Blue}{(e_N,h_1)(n_2,e_H)(e_N,h_1^{-1})}(e_N,h_1,)(e_N,h_2) \end{array} $$
and
$$\begin{array}{ll} (n_1,h_1)(n_2,h_2) & =(n_1\phi_{h_1}(n_2),h_1h_2) \\ & = (n_1,e_H)\color{Blue}{(\phi_{h_1}(n_2),e_H)}(e_N,h_1)(e_N,h_2). \end{array} $$
This means when we conjugate elements of $N\times\{e_H\}$ by elements of $\{e_N\}\times H$, we get the same thing as if we apply the elements of $H$ as automorphisms to $N$, then put it in $N\times\{e_H\}$.
Tuples are annoying and obfuscate the algebra in my opinion though. We should think of the semidirect product $N\rtimes H$ as the free product $N*H$ (whose elements are words formed from using the elements of $N$ and $H$ as letters) modulo the relation that conjugating elements of $N$ by elements of $H$ yields the same thing as if we applied the corresponding automorphism.
That is, elements of $N\rtimes H$ look like words formed from elements of $N$ and $H$. Their identity elements are identified as the same group element in $N\rtimes H$. Elements of $N$ multiply among themselves as usual, and same for elements of $H$ multiplying among themselves. But every instance of the word $hnh^{-1}$ ($h\in H,n\in N$) may be simplified to $\phi_h(n)$, and that is the only relation imposed on multiplication between elements of the two subgroups $N$ and $H$.
Using this definition, it's easy to see that $hn=(hnh^{-1})h=\phi_h(n)h$ so $HN=NH$ within $N\rtimes H$, and every element of $H$ can be "slid past" an element of $N$ to the right (although it changes the element of $N$ along the way). As a result, every word $\cdots h_{-1}n_{-1}h_0n_0h_1n_1\cdots$ (finitely many letters of course) can be simplified via this sliding rule to the canonical form $nh$.
Writing $n_1h_1=n_2h_2$ yields $h_1h_2^{-1}=n_1^{-1}n_2$, but the only element in $N\cap H$ (when we treat $N,H$ as subgroups of $N\rtimes H$) is the identity, so $h_1=h_2$ and $n_1=n_2$. Thus $N\rtimes H$ can be bijected with $N\times H$ set-theoretically. In order to transport the multiplication over, it remains to see how $(n_1h_1)(n_2h_2)$ simplifies to $n_3h_3$, which is something you've essentially already done.
