# Section5.2The subgroup lattices of cyclic groups¶ permalink

We now explore the subgroups of cyclic groups. A complete proof of the following theorem is provided on p. 61 of [1].

Sketch of proof: Let $G=\langle a\rangle$ and $H\leq G\text{.}$ If $H=\{e\}\text{,}$ then clearly $H$ is cyclic. Else, there exists an element $a^i$ in $H$ with $i>0\text{;}$ let $d$ be the least positive integer such that $a^d\in H\text{.}$ It turns out that $H=\langle a^d\rangle\text{.}$

Really, it suffices to study the subgroups of $\Z$ and $\Z_n$ to understand the subgroup lattice of every cyclic group.

We provide the following theorems without proof.

In fact:

##### Example5.2.6

How many subgroups does $\Z_{18}$ have? What are the generators of $\Z_{15}\text{?}$

Well, the positive divisors of $18$ are $1,2,3,6,9,$ and $18\text{,}$ so $\Z_{18}$ has exactly six subgroups (namely, $\langle 1\rangle\text{,}$ $\langle 2\rangle\text{,}$ etc.). The generators of $\Z_{15}$ are the elements of $\Z_{15}$ that are relatively prime to $15\text{,}$ namely $1,2,4,7,8,11,13,$ and $14\text{.}$

##### Example5.2.7

Draw a subgroup lattice for $\Z_{12}\text{.}$

The positive divisors of $12$ are $1,2,3,4,6,$ and $12\text{;}$ so $\Z_{12}$'s subgroups are of the form $\langle 1\rangle\text{,}$ $\langle 2\rangle\text{,}$ etc. So $\Z_{12}$ has the following subgroup lattice.