# How to show a group G with a given presentation is the semidirect product of two cyclic groups of prime order?

The question as posed in my homework assignment is as follows. I have been able to prove that the map $\sigma$ is an automorphism, and I think that the existence of the integer $t$ must be related to Fermat's Little Theorem, but after that I'm completely lost. Should I be demonstrating that $C_p \rtimes_{\phi} C_q$ has the same presentation as $G$?

Exercise: Let $G$ and $H$ be groups and let $\phi : H \to \text{Aut } (G)$ be a homomorphism. We define the semi-direct product of $G$ and $H$ (with respect to the homomorphism $\phi$) to be the cartesian product of $G$ and $H$ together with the following operation: \begin{align*} (g_1,h_1)(g_2,h_2) = (g_1[\phi(h_1)](g_2),h_1h_2). \end{align*} The semi-direct product of $G$ and $H$ with respect to $\phi$ is denoted $G \ltimes_{\phi} H$.

Let $p$ and $q$ be distinct primes with $p > q$ and $p \equiv 1 \mod q$. Let $G$ be the group determined by the generators $a$ and $b$ subject to the relations $|a| = p$, $|b| = q$, and $ba = a^t b$, where $t$ is a fixed integer satisfying $t^q \equiv 1 \mod p$, $t \not\equiv 1 \mod p$ (such an integer must exist - why?). Show that $G$ is a semi-direct product of the cyclic groups $C_p = \langle a \rangle$ and $C_q = \langle b \rangle$. [Hint: first show that the map $\sigma : C_p \to C_p$ given by $\sigma (a^j) = a^{jt}$ is an automorphism, and then consider the map $\phi : C_q \to \text{Aut } (C_p)$ given by $\phi (b^i) = \sigma ^i$.] The group $G$ is called a \textit{metacyclic} group.

For instance, you could take the following map as a candidate. Let $u, v$ be generators of $C_p$ and $C_q$ respectively. Define
$$f: C_p \rtimes C_q \to G$$ by $$(u^s, v^t) \mapsto a^sb^t.$$