Understanding the semidirect product I'm trying to motivate the definition of a semidirect product, and it seems like it comes from geometric examples. Some classic ones are
$D_{2n} \cong Z_n \rtimes Z_2$
$E(n) \cong R^2 \rtimes O(2)$
$O(n) \cong SO(n) \rtimes Z_2$
It seems like the group that is doing the acting sort of keeps track of the "orientation" of the group that is being acted upon. Can anyone provide a nice reference for understanding these examples, or an explanation?
 A: Let $G$ be a group and let $S$ be a torsor over $G$; this is a set on which $G$ acts in a way that is (non-canonically) isomorphic to the action of $G$ by left multiplication on itself, and is a natural generalization of an affine space. For example:


*

*$n$-dimensional Euclidean space is a torsor over $\mathbb{R}^n$.

*The set of total orders on a set of size $n$ is a torsor over the symmetric group $S_n$.

*The set of possible locations of a bug on some vertex of an $n$-gon is a torsor over the cyclic group $\mathbb{Z}/n\mathbb{Z}$. 


Then $S$ naturally inherits $G$ as a group of automorphisms (as a set). Now let $H$ be a subgroup of $\text{Aut}(G)$. Then picking an element $s \in S$ also allows us to identify $G$ with $S$ (via the map $g \mapsto gs$), hence defines an action of $H$ on $S$, but this action depends on the choice of $s$.
The subgroup of $\text{Aut}(S)$ (again, as a set) generated by $G$ and $H$ above is the semidirect product $G \rtimes H$ (regardless of the choice of $s$; exercise). Note the geometric significance of the case $G = \mathbb{R}^n, H = \text{O}(n)$. 
(Of course in general the homomorphism $H \to \text{Aut}(G)$ is not injective, but I think the geometric significance is clearest when it is. In general perhaps a more algebraic perspective is appropriate; look up "split exact sequence.") 
A: I think the best way to understand the semidirect product is not through geometry, but through pure algebra. Here is my point of view that I prefer (explained in detail in the Dummit & Foote). 
Given two subgroups of a group $G$ such that $H$ is normal in $G$, $HK = G$ and $H \cap K = \{1\}$, one can deduce that any element $g \in G$ has a unique expression in the form $g = hk$, where $h$ and $k$ are uniquely determined by $g$. Now if we try to multiply two elements of the group, then
$$
g_1 g_2 = (h_1 k_1)(h_2 k_2) = h_1 k_1 h_2 (k_1^{-1} k_1) k_2 = (h_1 (k_1 h_2 k_1^{-1})) (k_1 k_2) = h_3 k_3 
$$
In this manner, by unicity we have $k_1 k_2 = k_3$, but $h_3 = h_1 \varphi_{k_1}(h_2)$, where $\varphi_{k_1}$ gives you the way that "$K$ acts on $H$". Now $k_1$ is (except if it is the identity) not an element of $H$, so we cannot say that we obtain elements by conjugation in general ; when one constructs the semidirect product, the conjugation by an element of $K$ is replaced by automorphisms, and to mimic this case we use the abstract definition with the homomorphism $\varphi : K \to \mathrm{Aut}(H)$, but that's essentially what's behind the idea ; you get an "almost direct" product.
I hope that helps,
