5.2: The Subgroup Lattices of Cyclic Groups

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

Corollary \(\PageIndex{1}\)

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

Really, it suffices to study the subgroups of \(\mathbb{Z}\) and \(\mathbb{Z}_n\) to understand the subgroup lattice of every cyclic group.

We provide the following theorems without proof.

Theorem \(\PageIndex{2}\)

The nontrivial subgroups of \(\mathbb{Z}_n\) 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 \(d\text{.}\)

Theorem \(\PageIndex{3}\)

For each \(0\neq a\in \mathbb{Z}_n\text{,}\)

\begin{equation*} \langle a\rangle =\left\langle \dfrac{n}{\gcd(n,a)}\right\rangle. \end{equation*}

It follows that for each \(0\neq a \in \mathbb{Z}_n,\)

\begin{equation*} |\langle a\rangle |=\dfrac{n}{\gcd(n,a)}. \end{equation*}

In fact:

Theorem \(\PageIndex{4}\)

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

Example \(\PageIndex{1}\)

How many subgroups does \(\mathbb{Z}_{18}\) have? What are the generators of \(\mathbb{Z}_{15}\text{?}\)

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

Example \(\PageIndex{2}\)

Draw a subgroup lattice for \(\mathbb{Z}_{12}\text{.}\)

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