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I have solved several exercises on classifying groups and I have been wanting to generalize my results. however, I came across this problem where I know that there is no semi-direct products of $\mathbb{Z}_{p}$ by $\mathbb{Z}_{q}$, only direct products, since $p<q$, but, I had difficulty classifying the semi-direct products of $\mathbb{Z}_{q}$ by $\mathbb{Z}_{p}$.

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  • $\begingroup$ Just to be clear, you are looking for $\Bbb Z_q\rtimes \Bbb Z_p$? $\endgroup$ – Alex Ortiz Nov 5 '18 at 4:49
  • $\begingroup$ @AOrtiz exactly $\endgroup$ – Erick Vinicius Nov 5 '18 at 5:04
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We need to find all the homomorphisms $\Bbb Z_p\to \mathrm{Aut}(\Bbb Z_q)$. Recall that $\mathrm{Aut}(\Bbb Z_q) \cong \Bbb Z_{q-1}$ since $q$ is prime. Hence we need to know all the homomorphisms $\phi\colon \Bbb Z_p\to\Bbb Z_{q-1}$, each of which is determined by where we send the generator $1$ of $\Bbb Z_p$. For any such homomorphism, the order of $\phi(1)$ divides $p$ and it divides $q-1$. One option for $\phi$ is the trivial homomorphism, which corresponds to $1\mapsto \mathrm{Id}_{\Bbb Z_q}$, and that gives us the usual direct product $\Bbb Z_q\times\Bbb Z_p$.

The other option is $p$ divides $q-1$, i.e. where $q \equiv 1\bmod p$. Then the image $\phi(1)$ generates a unique cyclic subgroup of $\Bbb Z_{q-1}$ of order $p$, generated by $\frac{q-1}{p}$. This corresponds to the automorphism $\psi$ of $\Bbb Z_q$ defined by $\psi\colon 1\mapsto \frac{q-1}{p}$.

Hence the isomorphism classes of groups $G$ of order $pq$, with $p < q$ and $p,q$ primes are $$ G\cong \Bbb Z_q\times\Bbb Z_p\qquad\text{or}\qquad G\cong \Bbb Z_q\rtimes_\phi\Bbb Z_p, $$ where $\phi\colon \Bbb Z_p\to \mathrm{Aut}(\Bbb Z_q)$ is defined by $\phi\colon 1 \mapsto \psi$, where $\psi$ is the automorphism of $\Bbb Z_q$ we described above.

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