## Cyclic Groups

A *cyclic* group $G$ is a group that can be generated by
a single element $a$, so that every element in $G$ has the form $a^i$ for
some integer $i$.
We denote the cyclic group of order $n$ by $\mathbb{Z}_n$,
since the additive group of $\mathbb{Z}_n$ is a cyclic group of order $n$.

**Theorem:** All subgroups of a cyclic group are cyclic.
If $G = \langle a\rangle$ is cyclic, then for every divisor $d$ of $|G|$
there exists exactly one subgroup of order $d$ which may be generated
by $a^{|G|/d}$.

**Proof:** Let $|G| = d n$. Then
$1, a^n, a^{2 n},..., a^{(d-1)n}$ are distinct and form a cyclic subgroup
$\langle a^n \rangle$ of order $d$. Conversely,
let $H = \{1,a_1,...,a_{d-1}$ be a subgroup of $G$ for some $d$ dividing $G$.
Then for all $i$, $a_i = a^k$ for some $k$, and since every element has order
dividing $|H|$, $a_i^d = a^{k d} = 1$. Thus $k d = |G|m = n d m$ for some
$m$, and we have $a_i = a^{n m}$ so each $a_i$ is in fact a power
of $a^n$. From above this means it must be one of the $d$ subgroups already
described.

**Theorem:** Every group of composite order has proper subgroups.

**Proof:** Let $G$ be a group of composite order, and let $1\ne a\in G$.
Then if $\langle a \rangle \ne G$ we are done, otherwise
the subgroup $\langle a^d \rangle \ne G$ for every divisor $d$ of $|G|$.