Consider the cyclic group $\Bbb Z_n=\{1,2,\cdots n\}$ under addition modulo $n$ and for some non zero $a\in \{1,2,\cdots n-1\}$ let $\langle a\rangle=H\le \Bbb Z_n$ of order $q$. I wish to show that at most $\lceil q/2\rceil$ of the numbers in $H$ are in $\{1,2,\cdots \lceil n/2\rceil\}$ and the same holds for $\{\lceil n/2\rceil+1,\cdots n\}$.
Geometrically can I argue as follows: If $\Bbb Z_n$ is thought of as $n$th roots of unity lying on the unit circle then the points of $H$ starting from $1$ will be spaced by an angle of $2\pi/q$, i.e. will be equally spaced apart on the unit circle. So basically we have $q$ points equally spaced on a circle. Any diameter yields two segments each of which can contain at most $\lceil q/2\rceil$ of the points. A diameter contains $\lceil n/2 \rceil$ points of $\Bbb Z_n$.
I am not satisfied with the above argument (particularly the last two sentences). Can someone give a better proof?
Thanks.