Consider a group $G$ of order $pq$ (like ($\mathbb{Z}_p \times \mathbb{Z}_q$,+)), where $p$ and $q$ are distinct primes, and two elements $g_1, g_2 \in G$. Suppose $|g_1|=p$ and $|g_2|=q$. These two elements satisfy the relation,
$g_1^m g_2^n =e$, where $e$ is the identity element of $G$.
Then can we write as,
$m \equiv 0 (mod p)$
$n \equiv 0 (mod q)$ ?
If it is possible can someone please explain why it is possible? This is not clear to me..
Thanks a lot in advance.
There is a useful result in group theory: $\forall x\in G$, and $m, n\in \mathbb{Z}$, if $x^n = 1$ and $x^m = 1$, then $x^{(m, n)} = 1$ (where $(m, n)$ denotes the greatest common divisor of $m$ and $n$). To see why this holds, observe that $(m,n) = mx+ny$ for suitable $x$ and $y$ (see Extended Euclidean algorithm for how). Now from this it follows that $\forall m\in\mathbb{Z}, x^m = 1\iff |x|$ divides m. Since to say that $n$ divides $m$ is to say that $m\equiv 0(mod n)$, the rest follows.