5.2: The Subgroup Lattices of Cyclic Groups
- Page ID
- 84816
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)We now explore the subgroups of cyclic groups. A complete proof of the following theorem is provided on p. 61 of [1].
Theorem \(\PageIndex{1}\)
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 \(H=\langle a^d\rangle\text{.}\)
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.