Let $G=(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$, where $p$ and $q$ are odd distinct primes. Let $G$ be generated by the elements $s=(g_1,e_q)$ and $t=(e_p,g_3)$,
$g_1,e_p \in \mathbb{Z}_p \times \mathbb{Z}_p$,
$g_3,e_q \in \mathbb{Z}_q$,
$e_p,e_q$ are identity elements. $|s|=p, |t|=q$.
$\phi:\mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$
$e_q \rightarrow \phi_{e_q}$
$0 \rightarrow \phi_{0}$
$1 \rightarrow \phi_{1}$
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$q-1 \rightarrow \phi_{q-1}$
So for any $\phi_k, 0 \geq k \geq q-1$, we can determine $\phi_k(g_1)=g$, where $g \in \mathbb{Z}_p \times \mathbb{Z}_p$ ($g$ is some element and it will change according to $\phi_k$).
Let $\mathbb{Z}_p \times \mathbb{Z}_p$ be generated by $g_1$ and $g_2$. Let $|g_1|=|g_2|=p$. Then any $g \in \mathbb{Z}_p \times \mathbb{Z}_p$ can be expressed in terms of $g_1$ and $g_2$.
For, $\phi_k(g_1)=g$ let, $\phi_k(g_1)=g=g_1^{m_1} g_2^{m_2} \rightarrow (1)$
For a product of elements, say $sts$,
$(g_1,e_q)(e_p,g_3)(g_1,e_q)$ we can simplify as,
$(g_1 \phi_{e_q}(e_p), e_q g_3)(g_1,e_q)$
$(g_1 \phi_{e_q}(e_p) \phi_{g_3}(g_1), g_3 e_q)$
$(g_1 \phi_{e_q}(e_p) \phi_{g_3}(g_1), g_3) \rightarrow (2)$
Now as mentioned by (1), $\phi_{g_3}(g_1)$ also can be written in terms of $g_1, g_2$ and $\phi_{e_q}(e_p)=e_p$. Therefore, the first coordinate of the above ordered pair in (2) will simplify as some power of $g_1$ and $g_2$. Let it be as,
$(g_1^{m_3}g_2^{m_4}, g_3)$
Now suppose the group elements satisfies relationships such as,
$stst^{-1}s=e \rightarrow (3)$ (No. of elements in L.H.S. =5)
$ssts^{-1}t^{-1}st=e \rightarrow (4)$ (No. of elements in L.H.S. =7)
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$sts^{-1}ts...ts=e \rightarrow (5)$ (No. of elements in L.H.S. =$p^2q$)
Having $p^2q$ no. of elements in the L.H.S. is the maximum possible no. of elements that can be present such that the product is equal to the identity $e$, where $e=(e_p,e_q) \in (\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes_{\phi} \mathbb{Z}_q$.
In (3)-(5) also if we simplify in the way we obtained (2), we will get something similar.
Suppose we get as,
$(g_1^{m_5}g_2^{m_6}, g_3^{m_7})=(e_p,e_q)$ for (5). Can we solve for $m_5 , m_6 , m_7$ values if we know $p,q,g_1,g_2,g_3, \phi$? (May be using the data that (5) gives the maximum case?)
What is the method of obtaining the solutions? Can we solve for $m_5 , m_6 , m_7$ by a system of linear congruence relations?
Thanks a lot in advance.
Note: we don't know (5) but only knows that (5) gives maximum case and $p,q,g_1,g_2,g_3, \phi$ values. That's why I'm asking for a way of solving algebraically.
Even and idea or guidance regarding this is great. Many thanks again.