PreTeXt Sample Book Abstract Algebra (SAMPLE ONLY)

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let $G$ be a cyclic group generated by $a$ and suppose that $H$ is a subgroup of $G\text{.}$ If $H=\{e\}\text{,}$ then trivially $H$ is cyclic. Suppose that $H$ contains some other element $g$ distinct from the identity. Then $g$ can be written as $a^n$ for some integer $n\text{.}$ Since $H$ is a subgroup, $g^{-1}=a^n$ must also be in $H\text{.}$ Since either $n$ or $-n$ is positive, we can assume that $H$ contains positive powers of $a$ and $n>0\text{.}$ Let $m$ be the smallest natural number such that $a^m\in H\text{.}$ Such an $m$ exists by the Principle of Well-Ordering.

We wish to find the smallest integer $m$ such that $e=b^m=a^{km}\text{.}$ By Proposition 4.2.3, this is the smallest integer $m$ such that $n$ divides $km$ or, equivalently, $n/d$ divides $m(k/d)\text{.}$ Since $d$ is the greatest common divisor of $n$ and $k\text{,}$ $n/d$ and $k/d$ are relatively prime. Hence, for $n/d$ to divide $m(k/d)$ it must divide $m\text{.}$ The smallest such $m$ is $n/d\text{.}$