A group of order pq It is well-know that if $q\geq p>0$ be prime numbers, then G which is a group of order $pq$ is either isomorphic to $C_{pq}$ or $C_p \times C_q$ or $C_p \ltimes C_q$ and $p|q-1$.
I was wondering to know the difference of the structure of $C_p \times C_q$ and $C_p \ltimes C_q$. For example I know that $C_p \times C_q$ is abelian but I do not know why $C_p \ltimes C_q$ is not abelian and if $C_p \ltimes C_q$ is a simple group or not.
 A: The notation $C_q\rtimes C_p$ is of course ambiguous. What you need is a morphism $\phi\colon C_p\to \mathrm{Aut}(C_q)$, which is then used to define the semidirect product $C_q\rtimes_{\phi} C_p$ on the underlying set $C_q\times C_p$ by defining the product by the formula:
$$(a,b)(c,d) = (a\phi(b)(c),bd).$$
(Note: "$\phi(b)(c)$" looks weird, but makes sense: $b$ is an element of $C_p$, so $\phi(b)$ is an element of $\mathrm{Aut}(C_q)$, that is, an automorphism of $C_q$. That means it is a function that takes elements of $C_q$ as arguments, and returns elements of $C_q$; so we can evaluate the automorphism $\phi(b)$ at the element $c$, and we get the element $\phi(b)(c)$. )
If $\phi$ were, for example, the trivial map, so that $\phi(b)=\mathrm{id}_{C_q}$, this "semidirect product" would actually be a direct product. But the point of $p|q-1$ is that there exist a nontrivial map $\phi$ and this is used to define the semidirect product. (In fact, this would not generally suffice, as there may be several different nontrivial maps, but one can show that any two choices of nontrivial map will yield isomorphic groups).
So suppose that $\phi$ is nontrivial. That means that there exists $b\in C_p$ such that $\phi(b)$ is not the identity map, and therefore that there exists $c\in C_q$ such that $\phi(b)(c)\neq c$. Note however that $\phi(1)$ is always the identity map. Now consider the product of $(1,b)$ with $(c,1)$ in the two different orders. We have
$$\begin{align*}
(1,b)(c,1) &= (\phi(b)(c),b),\\
(c,1)(1,b) &= (c\phi(1)(1),b) = (c,b).
\end{align*}$$
But since $\phi(b)(c)\neq c$, then $(1,b)(c,1)\neq (c,1)(1,b)$. Thus, the group is not abelian. 
In fact, this argument shows that in general, if $G$ and $H$ are groups, and $\phi\colon G\to \mathrm{Aut}(H)$ is a nontrivial morphism, then the semidirect product $H\rtimes_{\phi}G$ cannot be abelian, regardless of whether $H$ and $G$ are abelian or not. 
The group is never simple, since it has a proper normal subgroup of order $q$.
