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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.

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    $\begingroup$ What is $g$ in $g^mg^n=e$? $\endgroup$
    – brj
    May 10, 2020 at 13:52
  • $\begingroup$ Sorry that was a mistake. Thanks @brj I corrected it now. $\endgroup$
    – Bob Traver
    May 10, 2020 at 17:24

1 Answer 1

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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.

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  • $\begingroup$ Thank you very much @brj . That means from $g_1^m g_2^n =e$, we can get $g_1^m=e \rightarrow (1)$ and $g_2^n=e \rightarrow (2)$. And then using (1) and $g_1^p=e$ we get $m \equiv 0 (mod p)$ and using (2) and $g_2^q=e$ we can get $n \equiv 0 (mod q)$ ? $\endgroup$
    – Bob Traver
    May 10, 2020 at 17:39
  • $\begingroup$ From (1) can we also take as $g_1^m=g_1^p$ and then get $m \equiv p (mod p)$ as well? And same for (2), as $n \equiv q (mod q)$? $\endgroup$
    – Bob Traver
    May 10, 2020 at 17:43
  • $\begingroup$ @BobTraver no, that is incorrect $g_1^mg_2^n$ does not imply $g_1^m$ and $g_2^n$ are $e$ (they can be inverses of each other) $\endgroup$
    – brj
    May 10, 2020 at 17:43
  • $\begingroup$ Ok, @brj but then in my question I know the relation $g_1^m g_2^n =e$ only, then how can I apply the above answer to get $m \equiv 0 (mod p)$ and $n \equiv 0 (mod q)$? $\endgroup$
    – Bob Traver
    May 10, 2020 at 17:54

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