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\(\def\Z{\mathbb{Z}} \def\zn{\mathbb{Z}_n} \def\znc{\mathbb{Z}_n^\times} \def\R{\mathbb{R}} \def\Q{\mathbb{Q}} \def\C{\mathbb{C}} \def\N{\mathbb{N}} \def\M{\mathbb{M}} \def\G{\mathcal{G}} \def\0{\mathbf 0} \def\Gdot{\langle G, \cdot\,\rangle} \def\phibar{\overline{\phi}} \DeclareMathOperator{\lcm}{lcm} \DeclareMathOperator{\Ker}{Ker} \def\siml{\sim_L} \def\simr{\sim_R} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} \)

Section5.2The subgroup lattices of cyclic groups

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:


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.

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