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This is exercise 7.29 in Rotman's book "An Introduction to the Theory of Groups".

If $p$ and $q$ are distinct primes, construct all semidirect products of $\mathbb{Z}_p$ by $\mathbb{Z}_q$.

I don't really understand the question. The group is of course $\mathbb{Z}_p \rtimes \mathbb{Z}_q$, but that is not an answer, it's really just restating the question. So what is the answer? Is it just stating that the group has order $ pq $ and that, for $ a,b\in \mathbb{Z}_p $ and $ x,y\in \mathbb{Z}_q $, $$ (a,x)(b,y)=(a\theta_x(b),xy)? $$ But again, I am not really doing anything but stating the obvious. What am I supposed to do? The exercise also ask me to compare the result with Theorem 4.20 about groups of order $ pq $. Am I supposed to find the identities like the ones in the theorem ($ b^p=1,a^q=1,aba^{-1}=b^m $)? If yes, how?

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  • $\begingroup$ I assume the point is to classify the possible $\theta$. $\endgroup$ – Mees de Vries Dec 5 '17 at 11:47
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    $\begingroup$ $\mathbb{Z}_p \ltimes \mathbb{Z}_q$ is not the group. It might be one of quite a few different groups. $\endgroup$ – Ivan Neretin Dec 5 '17 at 12:06
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To answer that question, one must analyse the homomorphisms $Z/q \to \operatorname{Aut}Z/p \cong Z/(p-1)$. There are exactly $\gcd(q,p-1)$ such homomorphisms; in particular, since $q$ is prime, there exist nontrivial homomorphisms iff $q$ divides $p-1$. Therefore:

If $q \not \mid p-1$ then $Z/p \rtimes Z/q \cong Z/p \times Z/q \cong Z/pq$.

Otherwise, pick a nontrivial homomorphism $\theta:Z/q \to \operatorname{Aut}Z/p$, say with $\theta_x := \theta(x)$ and $\theta_1(1) =: m \neq 1 \in Z/p$, and construct the semidirect product $G = Z/p \rtimes_\theta Z/q$. To compare with Theorem 4.20, let $a = (0,1) \in G$, $b = (1,0) \in G$. Then $G$ is generated by $a,b$, and we have the relations (written multiplicatively, as in the book) $b^p = 1$, $a^q = 1$, $aba^{-1} = \theta_a(b) = b^m$, and $m^q \equiv 1 \mod p$ but $m \not\equiv 1 \mod p$ as desired.

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  • $\begingroup$ 1. How can you define $\theta_x := \theta(x)$? I mean, in $\theta_x$, $x\in \mathbb{Z}_q$ but in $\theta(x)$, $x\in \mathbb{Z}_p$. 2. $\theta$ is a homomorphism, so how can you define $\theta_1(1) =: m \neq 1 \in \mathbb{Z}_p$? Shouldn't $\theta_1(1)=1$? 3, How do you get $\theta_a(b) = b^m$? $\endgroup$ – Barbara Dec 5 '17 at 13:38
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    $\begingroup$ @Barbara 1. I mean that for each $q \in Z/q$, we get an automorphism $\theta_q:Z/p \to Z/p$; this automorphism is the value of the homomorphism $\theta:Z/q \to \operatorname{Aut}Z/p$ at a a particular $q$. 2. The point is that a homomorphism from a cyclic group is uniquely determined by its value on the generator $1$. So the knowledge of $\theta_1$ entirely determines $\theta$, and the knowledge of $\theta_1(1)$ entirely determines $\theta_1$. This $\theta_1(1)$ need not be equal to $1$, however. (The groups are written in additive notation; so $\theta_1(0) = 0$ always but $\theta_1(1) = ?$.) $\endgroup$ – Alex Provost Dec 5 '17 at 17:17
  • $\begingroup$ @Barbara 3. The group $Z/p$ is embedded inside the semidirect product $G$ via the map $x \mapsto (x,0)$, and similarly for $Z/q$ with the other coordinate. So the generator $1 \in Z/p$ corresponds to $(1,0) = b \in G$, and the generator $1 \in Z/q$ corresponds to $(0,1) = a \in G$. In terms of the subgroups, this means that $\theta_a(b)$ is just computed as $\theta_1(1) = m \in Z/p$, and this corresponds to $(m,0) = b^m$ inside $G$. $\endgroup$ – Alex Provost Dec 5 '17 at 17:28
  • $\begingroup$ @Barbara I should have written $x \in Z/q$ and $\theta_x:Z/p \to Z/p$ in my first comment, sorry for the potentially confusing notation. $\endgroup$ – Alex Provost Dec 5 '17 at 17:36
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I should like to note that, in the case $\ell:=\frac{p-1}{q}$ is an integer, there are exactly two semidirect products of $\mathbb{Z}_p$ by $\mathbb{Z}_q$, up to isomorphism. One is the direct product $\mathbb{Z}_p \times \mathbb{Z}_q$. The other is given by $$H=\big\langle r,s\,\big|\,r^q=1\,,\,\,s^p=1\,\text{ and }rsr^{-1}=s^{g^\ell}\big\rangle\,,$$ where $g\in \mathbb{Z}_p$ is a fixed generator of the group $\left(\mathbb{Z}_p\right)^\times \cong \mathbb{Z}_{p-1}$. It can be easily seen that the group homomorphism from $G$ to $H$ induced by $a\mapsto s$ and $b\mapsto r^tsr^{-t}$, if $g^{t\ell}=m$, is an isomorphism of groups. (See the definitions of $G$, $a$, $b$, and $m$ in Alex Provost's answer above.)

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