Let $G$ be a cyclic group of order $n$. Prove that every subgroup $H$ of $G$ is of the form $<a^m>$ where $m$ is a divisor of $n$.
We have $o(a^m)=\frac{n}{gcd(m,n)}=\frac{n}{m}, ~~~(\because $ $m$ is a divisor of $n$)
That is order of a cyclic subgroup generated by $a^m$ equal to $o(<a^m>)=\frac{n}{m}.$ Here this cyclic subgroup $H$, say generated by $a^m$ is of the form $<a^m>$.
But how to show that "every subgroup (?) $H$ of $G$ is of the form $<a^m>$"
Assume the subgroup has two elements $a^{m_1}$ and $a^{m_2}$. Show that $a^{\gcd(m_1,m_2)}$ is also an element of that subgroup. Conclude that if the subgroup is generated by more than one element, then you can swap two of the generators for one.
Ok, first the definition of the cyclic group.
$\left(\exists \ a\in G\right)\left(\forall g\in G\right)\left(\exists k\in\mathbb{Z}\right)\ \ g = a^k$.
What we have to do is to show, that every subgroup of $G$ have that property. So let $H$ be any subgroup of $G$. From the definition every $h\in H$ can be written as $a^k$ for some $k\in\mathbb{Z}$. Maybe $a\in H$, maybe not - we don`t care.
Let $r = \min\left\lbrace k\in\mathbb{Z}: a^k\in H\right\rbrace$.
We have to show, that:
WLoG $a\not\in H$ (otherwise $H=G$, and we are done). Suppose, that $H\neq <a^r>$. It means, that there exist $h\in H$, witch can`t be written as $(a^r)^v$ for some $v\in\mathbb{Z}$. But $h\in G$, so $h = a^u$ for some $u\in\mathbb{Z}$, and $u = q\cdot r + b$, where $0<b<r$. But $a^b = a^u\cdot (a^r)^{-q} \in H$, so $b\ge r$. Contradiction.
