##### Theorem5.2.1

Every subgroup of a cyclic group is cyclic.

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

Every subgroup of a cyclic group is cyclic.

*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 \(H=\langle a^d\rangle\text{.}\)

Every subgroup of \(\Z\) is of the form \(n\Z\) for \(n\in \Z\text{.}\) (Note that \(n\Z\simeq \Z\) unless \(n=0\text{.}\))

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.

The nontrivial subgroups of \(\Z_n\) are exactly those of the form \(\langle d\rangle\text{,}\) where \(d\) is a positive divisor of \(n\text{.}\) Note that

\begin{equation*} |\langle d\rangle |=n/d \end{equation*}for each such \(d\text{.}\)

For each \(0\neq a\in \Z_n\text{,}\)

\begin{equation*} \langle a\rangle =\left\langle \frac{n}{\gcd(n,a)}\right\rangle. \end{equation*}It follows that for each \(0\neq a \in \Z_n,\)

\begin{equation*} |\langle a\rangle |=\frac{n}{\gcd(n,a)}. \end{equation*}In fact:

\(\Z_n\) has a unique subgroup of order \(k\text{,}\) for each positive divisor \(k\) of \(n\text{.}\)

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 \(14\text{.}\)

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.