Order of an element divides $m$ when $a^m \equiv 1 \pmod n$ https://brilliant.org/wiki/order-of-an-element/
I was referring to the above link for order of an element and in basic properties while proving property $1$ due to minimality of d, $d \le gcd(m,d)$ is written.
Is it because $mx+dy\ge d$ i.e. $gcd(m,d)\ge d$ but that inequality hold only for all positive $x$ and $y$ but there are cases when $x$ is positive and $y$ is negative, in that case $mx+dy \le d$ ie $gcd(m,d)\le d$?
Can someone help me what does the minimality of d actually mean and how did this inequality obtained $d \le gcd(m,d)$?

The order is defined to be the smallest positive $p$ such that $a^p \equiv 1 \pmod{n}$ holds.
Since we have shown that $a^{gcd(d.m)} \equiv 1 \pmod{n}$ and we know that $gcd(d,m)>0$, then we must have $d \le gcd(d,m)$.
