Pairs of generators in $C_{5} \times C_{5}\times C_{5}$

Question

I want to find which pairs of elements generate $C_{5} \times C_{5}\times C_{5}$. I have demonstrated that each element in this group generate the group with another 50 elements, but I want to find now how they are.

For example, if I have $a=(a, a^{2}, a^{3})\in C_{5} \times C_{5}\times C_{5}$, I would like to understand how I have to write $b\in C_{5} \times C_{5}\times C_{5}$ in such a way that $\langle a, b\rangle = C_{5} \times C_{5}\times C_{5}$.

Any idea? thanks.

Answer 1

Firstly $a=(a,a^2,a^3)$ is nonsense, I assume you mean $a=(g,g^2,g^3)$ where $g$ generates $C_5$.

There are no such pairs.

Hint of proof:

Fix $a,b\in C_5\times C_5\times C_5$. What are the orders of $\langle a\rangle$ and $\langle b\rangle$?

We can write $\langle a,b\rangle=\{a^i,b^j|i,j\in\{0,1,2,3,4\}\}$ (why?) so the order of $\langle a,b\rangle$ is...