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. $|s|=p, |t|=q$.

$\phi:\mathbb{Z}_q \rightarrow Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$

$e_q \rightarrow \phi_{e_q}$

$0 \rightarrow \phi_{0}$

$1 \rightarrow \phi_{1}$




$q-1 \rightarrow \phi_{q-1}$

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)




$sts^{-1}ts...ts=e \rightarrow (5)$ (No. of elements in L.H.S. =$p^2q$)

(These relationships correspond to finding cycles in Cayley graphs if we draw the Cayley graph of the group $G$ with repect to $s,t$.)

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.

  • 1
    $\begingroup$ Assuming you know p,q, then (g1^a g2^b, g3^c) = (g1^d g2^e, g3^f) if and only if (1) a-d is divisible by p, (2) b-e is divisible by p, and (3) c-f is divisible by q. Therefore these numbers are usually reduced mod p and q, so that m5=0, m6=0, m7=0 after reduction. You may find the setup easier to work with if you replace Zp x Zp with GF(p^2) and Zq with a cyclic subgroup of the multiplicative group of GF(p^2) of order q. (In other words, choose an irreducible quadratic divisor of the polynomial x^q -1 over GF(p).) $\endgroup$ – Jack Schmidt Jun 17 '20 at 13:22
  • $\begingroup$ Thank you very very much @JackSchmidt is it possible to connect it with the maximum nature of the case as well by considering GF(p^2) ? $\endgroup$ – Buddhini Angelika Jun 20 '20 at 9:05

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