Definition 30. Cyclic Subgroup.
Let \(G\) be a group. The cyclic subgroup generated by an element \(g\in G\) is the set \(\subgroup{ g } = \{ g^n \mid n \in \Z\}\) (where \(g^0:=e\)).
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Section Cyclic Groups
Worksheet Part 1: Cyclic Subgroup Computations
The goal of these problems is to expand your familiarity with computations in cyclic groups and with cyclic subgroups.
1.
Convince yourself that \(\subgroup{g}\) is actually a subgroup.
2.
Let \(G\) be a cyclic group of order 30 generated by \(a\) (for instance, \(G\) could be \(\Z_{30}\) with generator \(7\)). Compute each of the following subgroups.
(a)
\(\subgroup{ a^4}\)
(b)
\(\subgroup{ a^{26}}\)
(c)
\(\subgroup{ a^6}\)
(d)
\(\subgroup{ a^{24}}\)
(e)
\(\subgroup{ a^3}\)
(f)
\(\subgroup{ a^{21}}\)
3.
How are these subgroups related to each other? Can you see any patterns in the relationships of the powers of \(a\) on the generators to each other and/or the order of the group \(n=30\text{?}\)
Worksheet Part 2: Generators for Cyclic Groups
In this part of the activity, you’ll think about generators for cyclic subgroups in general based on your work in the first part and practice filling in the details of a proof.
Theorem 31. Subgroups Generated by Powers of an Element.
Let \(a\) be an element of order \(n\) in a group and let \(k\) be a positive integer. Then \(\subgroup{ a^k} =\subgroup{ a^{\gcd(n,k)}}\) and \(|a^k|=n/\gcd(n,k)\text{.}\)
A partial proof of this theorem is given below. Fill in the missing details.
Proof.
Let \(d=\gcd(n,k)\) and let \(k=dr\text{.}\) Since \(a^k = (\fillinmath{XXX})^r\) we have by closure that \(\fillinmath{XXXX}\subseteq \subgroup{ a^d}\text{.}\) By Bezout’s lemma, there are integers \(s\) and \(t\) such that \(d=ns+kt\text{.}\) So
\begin{equation*}
a^d= \fillinmath{XXX} = \fillinmath{XXX} = e(a^k)^t = (a^k)^t \in \fillinmath{XXX}.
\end{equation*}
This proves . So we have verified that \(\subgroup{ a^k} =\subgroup{ a^{\gcd(n,k)}}\text{.}\)
To prove the second statement, we will show that \(|a^{d'}|=n/d'\) for any divisor \(d'\) of \(n\text{.}\) Certainly
\begin{equation*}
(a^{d'})^{\fillinmath{XX}}=a^n=e\text{,}
\end{equation*}
so \(|a^{d'}|\leq \fillinmath{XXX}\text{.}\) On the other hand, if \(i\) is a positive integer with \(i\lt n/d'\) then \(\fillinmath{XXXXXX}\) because \(d'i\lt n\) and the definition of \(|a|\text{.}\) Therefore we have \(|a^{d'}| \geq \fillinmath{XXX}\) and thus \(\fillinmath{XXXXXX}\text{.}\)
We now apply this fact with \(d'=\fillinmath{XXXX}\) to obtain
\begin{equation*}
|a^k|=|\subgroup{ a^k}|=|\subgroup{ a^{\fillinmath{XX}}}|=|a^{\fillinmath{XX}}|=\fillinmath{XXX}\text{.}
\end{equation*}
Solution to Proof Exercise.
Let \(d=\gcd(n,k)\) and let \(k=dr\text{.}\) Since \(a^k = (\mathbf{a^d})^r\) we have by closure that \(\mathbf{\subgroup{a^k}}\subseteq \subgroup{ a^d}\text{.}\) By Bezout’s lemma, there are integers \(s\) and \(t\) such that \(d=ns+kt\text{.}\) So
\begin{equation*}
a^d= \mathbf{a^{ns+kt}} = \mathbf{a^{ns}a^{kt}} = e(a^k)^t = (a^k)^t \in \mathbf{\subgroup{a^d}}.
\end{equation*}
This proves \(\mathbf{\subgroup{a^d}\subseteq \subgroup{a^k}}\text{.}\) So we have verified that \(\subgroup{ a^k} =\subgroup{ a^{\gcd(n,k)}}\text{.}\)
To prove the second statement, we will show that \(|a^{d'}|=n/d'\) for any divisor \(d'\) of \(n\text{.}\) Certainly
\begin{equation*}
(a^{d'})^{\mathbf{n/d'}}=a^n=e\text{,}
\end{equation*}
so \(|a^{d'}|\leq \mathbf{n/d'}\text{.}\) On the other hand, if \(i\) is a positive integer with \(i\lt n/d'\) then \(\mathbf{a^i\neq e}\) because \(d'i\lt n\) and the definition of \(|a|\text{.}\) Therefore we have \(|a^{d'}| \geq \mathbf{n/d'}\) and thus \(\mathbf{|a^{d'}|=n/d'}\text{.}\)
We now apply this fact with \(d'=\mathbf{\gcd(n,k)}\) to obtain
\begin{equation*}
|a^k|=|\subgroup{ a^k}|=|\subgroup{ a^{\mathbf{\gcd(n,k)}}}|=|a^{\mathbf{\gcd(n,k)}}|=\mathbf{n/\gcd(n,k)}\text{.}
\end{equation*}
Corollary 32. Criterion for Equality of Cyclic Subgroups.
Let \(|a|=n.\) Then \(\subgroup{ a^i}=\subgroup{ a^j}\) if and only if \(\gcd(n,i)=\gcd(n,j)\) and \(|a^i|=|a^j|\) if and only if \(\gcd(n,i)=\gcd(n,j)\text{.}\)
1.
Convince yourself why Criterion for Equality of Cyclic Subgroups follows from Subgroups Generated by Powers of an Element.
2.
In the group \(\Z_{30}\text{,}\) determine all of the subgroups \(\subgroup{ i}\) that are equal to each subgroup given below.
(a)
\(\subgroup{ 2}\)
(b)
\(\subgroup{ 7 }\)
(c)
\(\subgroup{ 12}\)
