Originally Posted by
DrSteve Let G be a cyclic group, say $\displaystyle G = <g>$, and $\displaystyle H\leq G$. Since $\displaystyle H\leq G$, all elements of $\displaystyle H$ are of the form $\displaystyle g^k$ for some integer $\displaystyle k$. If $\displaystyle H = \{ e\}$, then $\displaystyle H$ is cyclic, so assume $\displaystyle \{ e\} < H$ and let $\displaystyle k$ be the least positive integer such that $\displaystyle g^k$ is in $\displaystyle H$. Let $\displaystyle g^s$ be any other element of $\displaystyle H$. By the division algorithm we can write $\displaystyle s = qk + r$ with $\displaystyle 0\leq r < k$. $\displaystyle g^r = g^{s-qk} = g^s(g^{-k})^q$ is in $\displaystyle H$ since $\displaystyle g^k$ and $\displaystyle g^s$ are. By the definition of $\displaystyle k$, $\displaystyle r$ must be $\displaystyle 0$. So $\displaystyle g^s = (g^k)^q$. and $\displaystyle H = <g^k>$.