Cyclic groups

Section 1.9 Cyclic groups

Our approach to studying cyclic groups will make heavy use of some elementary number theory concepts, most of which we haven’t officially proved in this course yet. You can rest assured that we will eventually cover these results (specifically, when treading principal ideal domains in Math 331-2). We get ahead of ourselves by introducing them now mainly for the convenience of using these techniques in computational examples. For example, Theorem 1.9.1 provides an easy method for computing greatest common divisors and least common multiples from the prime factorizations of two given integers.

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Theorem 1.9.1. Prime factorization formulas.

Let \(m\) and \(n\) be positive integers, and let

\begin{align*} m= \amp \prod_{i=1}^rp_i^{m_i} \amp n\amp=\prod_{i=1}^rp_i^{n_i}\text{,} \end{align*}

where \(\{p_1,p_2,\dots, p_r\}\) is a list of distinct prime integers, and \(m_i, n_i\in \Z_{\geq 0}\) for all \(1\leq i\leq r\text{.}\)

We have

\begin{align} \gcd(m,n) \amp =\prod_{i=1}^rp_i^{\min\{m_i,n_i\}} \amp \lcm(m,n)\amp=\prod_{i=1}^r p_i^{\max\{m_i, n_i\}}\text{.}\tag{1.9.1} \end{align}

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We have

\begin{equation} mn=\gcd(m,n)\lcm(m,n)\text{.}\tag{1.9.2} \end{equation}

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Proposition 1.9.2. Order of group elements.

Let \(g\) be an element of \(G\text{.}\)

If \(\ord g=\infty\text{,}\) then \(g^i=g^j\) if and only if \(i=j\text{.}\)
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If \(\ord g=n\) for \(n\in \Z\text{,}\) then \(g^i=g^j\) if and only if \(i\equiv j \pmod n\text{.}\)
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If \(g^i=g^j=e\text{,}\) for integers \(i,j\in \Z\text{,}\) then \(g^{\gcd(i,j)}=e\text{.}\)
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If \(\ord g=n\text{,}\) then

\begin{equation} \ord g^i=\lcm(i,n)=\frac{n}{\gcd(i,n)}\tag{1.9.3} \end{equation}

for all \(i\in \Z\text{.}\)

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Theorem 1.9.3. Isomorphic cyclic groups.

Let \(G=\angvec{g}\) and \(H=\angvec{h}\) be cyclic groups.

\(\abs{G}=\ord g\text{.}\)
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\(G\cong H\) if and only if \(\abs{G}=\abs{H}\text{.}\)
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Proof.

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Corollary 1.9.4. Isomorphic cyclic groups.

If \(G\) is cyclic, then \(G\cong \Z\) or \(G\cong \Z/m\Z\) for some positive integer \(m\text{.}\)

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Theorem 1.9.5. Cyclic group equivalence.

A group \(G\) is cyclic if and only if all subgroups of \(G\) are cyclic.

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Proof.

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Theorem 1.9.6. Subgroups of infinite cyclic group.

Let \(G=\angvec{g}\) and assume \(\abs{G}=\infty\text{.}\)

All subgroups of \(G\) are cyclic.
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The correspondence

\begin{align*} \Z_{\geq 0} \amp \rightarrow \{H\subseteq G\mid H \text{ a subgroup}\}\\ m \amp \longmapsto H_m=\angvec{g^m} \end{align*}

is a bijection. In other words, every subgroup of \(G\) is equal to exactly one of the following list of subgroups:

\begin{equation*} H_0=\angvec{e}, H_1=\angvec{g}, H_2=\angvec{g^2},\dots \text{.} \end{equation*}

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Proof.

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Theorem 1.9.7. Subgroups of finite cyclic group.

Let \(G=\angvec{g}\) and assume \(\abs{G}=n< \infty\text{.}\)

All subgroups of \(G\) are cyclic.
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The correspondence

\begin{align*} \{d\in \{1,2,\dots, n\}\mid d\mid n\} \amp \rightarrow \{H\subseteq G\mid H \text{ a subgroup}\} \\ d \amp \mapsto H_d=\angvec{g^d} \end{align*}

from the positive divisors of \(n\) and the subgroups of \(G\text{.}\) Furthermore, \(\abs{H_d}=n/d\text{.}\) As a result for each positive divisor of \(n\) there is a unique subgroup of that cardinality.

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Proof.

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