Classify all groups with order $pq$, $p$ is not equal to $q$ I am reading the following material:
http://www.math.ucla.edu/~iacoley/hw/algqual.pdf
. At page 5, there is a proof classifying all groups with order $pq$(they are all primes and $p$ is not equal to $q$). I am lost in the last two sentence, which claims the semidirect product are all isomorphic. Could someone help me to understand it? Thanks!
 A: I'll try my best to explain both sentences in turn. Apologies in advance if I get anything wrong.

For two non-trivial homomorphisms $\varphi, \psi : P \rightarrow \text{Aut}Q$, we can construct an isomorphism of
  $Q \rtimes_\varphi P$ and $Q \rtimes_\psi P$ via an automorphism of $P$, since $P$ is cyclic.

Since $P$ has a prime order $p$, all of its non-trivial elements are generators. This gives us precisely $p - 1$ unique automorphisms of $P$, since what matters is where the automorphism sends the generators. Specifically, one particular generator can be sent to any of the $p - 1$ generators in $P$.
Take a generator $g \in P$. Then, we know $\varphi(g) = a_\varphi$ and $\psi(g) = a_\psi$ for $a_\varphi, a_\psi \in \text{Aut}Q$. If $a_\phi = a_\psi$, then things are straightforward, so let's suppose $a_\phi \neq a_\psi$. Then, we take an automorphism $\alpha : P \rightarrow P$ such that $\psi(\alpha(g)) = a_\varphi = \varphi(g)$.
We are guaranteed such an $\alpha$ exists since the images of $\varphi$ and $\psi$ are the same unique subgroup of $\text{Aut}Q$ of order $p$, which I will call $\text{Aut}Q^*$. Since $|P| = |\text{Aut}Q^*|$, $\varphi$ and $\psi$ are surjective onto $\text{Aut}Q^*$, and so for any $a \in \text{Aut}Q^*$, there exist $g,g' \in P$ such that $\varphi(g) = a$ and $\psi(g') = a$. We are simply finding the automorphism $\alpha$ where $\alpha(g) = g'$, and thus $\psi(\alpha) = \varphi$.
Take a mapping $\beta : Q \rtimes_\varphi P \rightarrow Q \rtimes_\psi P$ given by:
$$\beta : (q,p) \rightarrow (q,\alpha(p))$$
Suppose $\beta((q_1,p_1)) = \beta((q_2,p_2))$. Then, $(q_1,\alpha(p_1)) = (q_2,\alpha(p_2))$, which gives us $q_1 = q_2$ and $\alpha(p_1) = \alpha(p_2)$; apply the inverse of $\alpha$ to obtain $p_1 = p_2$, so $\beta$ is injective.
Take $(q,p) \in Q \rtimes_\psi P$. Then, we clearly have $\beta((q,\alpha^{-1}(p))) = (q,p)$, so $\beta$ is surjective (and therefore bijective). Furthermore, we have:
$$\begin{equation}
\begin{split}
\beta((q_1,p_1) \times_\varphi (q_2,p_2)) & = \beta((q_1\varphi(p_1)(q_2),p_1p_2)) \\ & = (q_1\varphi(p_1)(q_2),\alpha(p_1p_2)) \\ & = (q_1\psi(\alpha(p_1))(q_2),\alpha(p_1)\alpha(p_2)) \\ & = (q_1,\alpha(p_1)) \times_\psi (q_2,\alpha(p_2))) \\ & = \beta((q_1,p_1)) \times_\psi\beta((q_2,p_2))
\end{split}
\end{equation}$$
where $\times_\psi$ and $\times_\varphi$ are the operations of $Q \rtimes_\psi P$ and $Q \rtimes_\varphi P$, respectively; thus, $\beta$ is a homomorphism. Therefore, since $\beta$ is bijective, it is also an isomorphism, and so $Q \rtimes_\varphi P \cong Q \rtimes_\psi P$. This applies to all such pairs of semidirect products with nontrivial homomorphisms $\varphi, \psi$, so there is only one (up to isomorphism) such semidirect product.

Hence there are two nonisomorphic groups of order $pq$, which correspond to a trivial $\varphi$ and a nontrivial $\varphi$.

Combined with what was found earlier in the solution, we have two total cases:


*

*If $p \nmid q - 1$, then $G \cong \mathbb{Z}_{pq}$.

*If $p \mid q - 1$, then $G \cong Q \rtimes_\varphi P$ for a normal $q$-Sylow subgroup $Q$, a $p$-Sylow subgroup $P$, and a homomorphism $\varphi : P \rightarrow \text{Aut}Q$.


Thus, we have classified groups of order $pq$ into two nonisomorphic groups up to isomorphism.
