TITLE: Intuition about the semidirect product of groups
QUESTION [33 upvotes]: If we have two groups $G,H$ the construction of the direct product is quite natural. If we think about the most natural way to make the Cartesian product $G\times H$ into a group it is certainly by defining the multiplication
$$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2),$$
with identity $(1,1)$ and inverse $(g,h)^{-1}=(g^{-1},h^{-1})$.
On the other hand we have the construction of the semidirect product which is as follows: consider $G$,$H$ groups and $\varphi : G\to \operatorname{Aut}(H)$ a homomorphism, we define the semidirect product group as the Cartesian product $G\times H$ together with the operation
$$(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1\varphi(g_1)(h_2)),$$
and we denote the resulting group as $G\ltimes H$.
We then show that this is a group and show many properties of it. My point here is the intuition.
This construction doesn't seem quite natural to make. There are many operations to turn the Cartesian product into a group. The one used when defining the direct product is the most natural. Now, why do we give special importance to this one?
What is the intuition behind this construction? What are we achieving here and why this particular way of making the Cartesian product into a group is important?

REPLY [3 votes]: Many have given good answers here, so I just want to answer specifically for the intuition behind it.
Semi direct product came into light when we found out that if a group $H$ is a normal subgroup, and another group $K$ is also a subgroup (not necessarily normal) of a bigger group, and $H \cap K = 1$, the multiplication of those two subgroups would yield another group $HK$ with order $\frac{|H||K|}{|H \cap K|} = |H||K|$ (as $H \cap K = 1$).
So we know that the new group $HK$ can be written uniquely in the form of $hk$ where $h \in H$ and $k \in K$. Because of that, the multiplication in $HK$, thus all elements of $HK$, must can always be written in the form of well... $hk$, e.g.:
$(h_1 k_1)(h_2 k_2) = h_1 k_1 h_2 (k_1^{-1} k_1) k_2 = h_1 (k_1 h_2 k_1^{-1}) k_1 k_2$
As $H$ is a normal subgroup, $k_1 h_2 k_1^{-1} \in H$, so it can be re-written as:
$(h_1 k_1)(h_2 k_2) = (h_1 (k_1 h_2 k_1^{-1})) (k_1 k_2) = h_3 k_3$
where $h_3 = h_1 (k_1 h_2 k_1^{-1}) \in H$ and $k_3 = k_1 k_2 \in K$.
But we notice that this left conjugation by $k_1$ which is $k_1 h_2 k_1^{-1}$ is an automorphism of $H$, so of course any automorphism of $H$ would do the job. If we define a homomorphism:
$\varphi: K \rightarrow Aut(H)$,
that homomorphism can be used in the place of left conjugation by $k_1$ and again achieve the same form of $hk$. Rewriting the above derivation with direct product notation and the defined homomorphism, we would get:
$(h_1, k_1)(h_2, k_2) = (h_1 \: \varphi(k_1)(h_2), k_1 k_2) = (h_3, k_3)$
where $h_3 = h_1 \: \varphi(k_1)(h_2) \in H$ and $k_3 = k_1 k_2 \in K$.
which is exactly the definition of semi direct product multiplication you ask.