Product of two elements in a semidirect product with distinct prime powers

Recall the definition of semidirect product: Let $G, H$ be groups and $\phi:H\longrightarrow \text{Aut}(G)$ a group homomorphism. We define the semidirect product of $G$ and $H$ ($G\rtimes_\phi H$) with multiplication given by $(g_1,h_1)(g_2,h_2)=(g_1\phi(h_1)(g_2),h_1h_2)$.

Question: Let $A = \mathbb{Z}_q \rtimes_{\phi} \mathbb{Z}_p$, such that $q \equiv 1 \mod p$. Where $q,p$ are prime. Where $\psi: \mathbb{Z}_q \rightarrow \mathbb{Z}_q$ is a group automorphism. Where $\phi: \mathbb{Z}_p \rightarrow \psi(\mathbb{Z}_p)$ is a (nontrivial, as noted by Derek Holt) group homomorphism. Let $x,y \in A$ such that $|x| = q, |y| = p$, show that $|xy| = p$.

Note as well that the Sylow theorems tells us that there is 1 element of order 1 (namely the identity), $q - 1$ elements of order $q$, and $q(p-1)$ elements of order $p$.

An example of this is consider when $q=7,p=3$. So $A = \mathbb{Z}_7 \rtimes_\phi \mathbb{Z}_3$. Let $\mathbb{Z}_7 := \lbrace 1,x,x^2,x^3,x^4,x^5,x^6 \rbrace$ and $\mathbb{Z}_3 := \lbrace 1,y,y^2 \rbrace.$ We define $\psi: \mathbb{Z}_7 \rightarrow \mathbb{Z}_7$ a group automorphism. Since the order of $\mathbb{Z}_3$ is 3 an example of $\psi$ is: \begin{array}{|c|c|c|} \hline x& \psi(x) & \psi^2(x) \\ \hline 1& 1& 1\\ \hline x& x^2&x^4 \\ \hline x^2& x^4&x^8=x \\ \hline x^3& x^6&x^{12}=x^5 \\ \hline x^4& x^8=x&x^2 \\ \hline x^5& x^{10}=x^3&x^6 \\ \hline x^6& x^{12}=x^5 &x^{10}=x^3 \\ \hline \end{array} and $1_{\mathbb{Z}_7}$ which sends any element of $\mathbb{Z}_7$ to itself. We define $\phi:\mathbb{Z}_3 \rightarrow \psi(\mathbb{Z}_7)$ a group homomorphism. An example of $\phi$ is
\begin{array}{|c|c|} \hline y& \phi(y)\\ \hline 1& 1_{\mathbb{Z}_7}\\ \hline y& \psi \\ \hline y^2& \psi^2\\ \hline \end{array} So an element in $A$ looks like $(1,1),(1,y),(1,y^2),(x,1),(x,y),(x,y^2),...,(x^6,1),(x^6,y),(x^6,y^2)$. An example of a product of two elements in $A$ is \begin{align*} (x^3,y^2)(x^5,y^2)&=(x^3\phi(y^2)(x^5),y^2y^2)\\ &= (x^3\psi^2(x^5),y^4)\\ &= (x^3x^6,y)\\ &= (x^9,y)\\ &=(x^2,y) \end{align*}

Really having a hard time showing this, I need to show this since I'm trying to show that the davenport constant of $A$ is $q + p -1$. Would really help if anyone has any idea on how to start or guide me through this!

• You have not said what $\phi$ is in your definition of $A$. You need to specify that $\phi$ is nontrivial. As for your problem, there are only $q-1$ elements of order $q$, namely the nontrivial powers of $x$, and $xy$ is not a power of $x$, so it must have order $p$. – Derek Holt Apr 10 '18 at 14:24