# Finding the relationship given by the generating elements of a non-abelian group (for a maximum case) where they correspond to finding cycles

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.

• 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).) – Jack Schmidt Jun 17 '20 at 13:22
• 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) ? – Buddhini Angelika Jun 20 '20 at 9:05