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 $n$.

Question

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 $<a^m>$"

Answer 1

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:

1. $n = r\cdot l$, for some $l\in\mathbb{N}$,
2. $H = a^r\cdot G = <a^r>$.

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.

Answer 2

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.