2.4: Introduction to Cyclic groups

Definition: order of an element

Let \((G,\star)\) be a group.

Let \(a \in G\), then the order of the element \(a \in G\), be the smallest positive integer \(n\) s.t. \(a^n=e\).  If no such \(n\) exists, we say that the order is infinite.

 

Definition: 

Let \((G,\star)\) be a group.

Let \(a \in G\), then define \(<a>=\{a^m|m\in \mathbb{Z}\}\).

Theorem \(\PageIndex{1}\)

 

Let \((G,\star)\) be a group.

Let \(a \in G\), then

\(<a>  \leq G.\)

Example \(\PageIndex{1}\)

Consider the group \(\mathbb{Z}_{10}\) with addition modulo 10.  What is the order of its elements?

Consider that \(\mathbb{Z}_{10}=\{0,1,2,3,4,5,6,7,8,9\}\).   Note that \(\{0\}\) is the identity and its order is \(1. \) 

element	Order	Calculations
0	\(|<0>|=1\)
1	\(|<1>|=10\)	\(1^{10}\equiv 0 \pmod{10}\)
2	\(|<2>|=5\)	\(2^{1}\equiv 2\pmod{10}\), \(2^2 \equiv 4 \pmod{10}\), \(2^{3}\equiv 6 \pmod{10}\), \(2^{4} \equiv 8 \pmod{10}\), and \(2^{5}\equiv 0 \pmod{10}\).
3	\(|<3>|=10\)	\(3^1\equiv 3 \pmod{10}\), \(3^2\equiv 6 \pmod{10}\) , \(3^3\equiv 9 \pmod{10}\), \(3^4\equiv 2 \pmod{10}\), \(3^5\equiv 5 \pmod{10}\), \(3^6\equiv 8 \pmod{10}\), \(3^7\equiv 1 \pmod{10}\), \(3^8\equiv 4 \pmod{10}\), \(3^9\equiv 7 \pmod{10}\), and  \(3^{10}\equiv 0 \pmod{10}\).
4	\(|<4>|=5\)	\(4^1 \equiv 4 \pmod{10}\), \(4^2 \equiv 8 \pmod{10}\), \(4^3 \equiv 2 \pmod{10}\), \(4^4 \equiv 6 \pmod{10}\), and \(4^5 \equiv 0 \pmod{10}\).
5	\(|<5>|=2\)	\(5^1 \equiv 5 \pmod{10}\), and \(5^2 \equiv 0 \pmod{10}\).
6	\(|<6>|=5\)	\(6^1 \equiv 6 \pmod{10}\), \(6^2 \equiv 2 \pmod{10}\), \(6^3 \equiv 8 \pmod{10}\), \(6^4 \equiv 4 \pmod{10}\) and \(6^5 \equiv 0 \pmod{10}\).
7	\(|<7>|=10\)	\(7^1 \equiv 7 \pmod{10}\), \(7^2 \equiv 4 \pmod{10}\), \(7^3 \equiv 1 \pmod{10}\), \(7^4 \equiv 8 \pmod{10}\), \(7^5 \equiv 5 \pmod{10}\), \(7^6 \equiv 2 \pmod{10}\), \(7^7 \equiv 9 \pmod{10}\), \(7^8 \equiv 6 \pmod{10}\), \(7^9 \equiv 3 \pmod{10}\), and \(7^{10} \equiv 0 \pmod{10}\).
8	\(|<8>|=5\)	\(8^1 \equiv 8 \pmod{10}\), \(8^2 \equiv 6 \pmod{10}\), \(8^3 \equiv 4 \pmod{10}\), \(8^4 \equiv 2 \pmod{10}\), and  \(8^5 \equiv 0 \pmod{10}\).
9	\(|<9>|=10\)	\(9^1 \equiv 9 \pmod{10}\), \(9^2 \equiv 8 \pmod{10}\), \(9^3 \equiv 7 \pmod{10}\), \(9^4 \equiv 6 \pmod{10}\), \(9^5 \equiv 5 \pmod{10}\), \(9^6 \equiv 4 \pmod{10}\), \(9^7 \equiv 3 \pmod{10}\), \(9^8 \equiv 2 \pmod{10}\), \(9^9 \equiv 1 \pmod{10}\),  and \(9^{10} \equiv 0 \pmod{10}\).

 

 

Note

Let \((G,\star)\) be a group.

Let \(a \in G\), then \(<a> \) is called cyclic subgroup of \( G.\)

Definition: Cyclic subgroup

If  \( G=<a>\)  for some \(a \in G\), then \(G\) is called a cyclic group.

Example \(\PageIndex{2}\)

\(U(15)=\{1,2,4,7,8,11,13,14\}\) and \(\star= \bullet \pmod{15}\).

Let \(a \in U(15)\) and \(b \in U(15)\).

Then \(a(x)+15(y)=1\) and \(ab(x)+15by=b\)  therefore \(U(15)\) is closed.

Since \(U(15)\) is a finite group, the order of the group is \(|U(15)|=8\).

However, \(U(15)\) is not a cyclic group. 

Proof: (by exhaustion)

Let \(a \in U(15)\).

We will show that \(\not \exists a \in U(15)\), s.t. \(|<a>|=8\).

\(|<1>|=1\) since \(1^1 \equiv 1 \pmod{15}\)

\(|<2>|=4\) since \(2^4 \equiv 1 \pmod{15}\)

\(|<4>|=2\) since \(4^2 \equiv 1 \pmod{15}\)

\(|<7>|=4\) since \(7^4 \equiv 1 \pmod{15}\).

  \(|<8>|=4\) since \(8^4 \equiv 1 \pmod{15}\).

\(|<11>|=2\) since \(11^2 \equiv 1 \pmod{15}\).

\(|<13>|=4\) since \(13^4 \equiv 1 \pmod{15}\).

\(|<14>|=2\) since \(14^2 \equiv 1 \pmod{15}\).

Since \(\not \exists a \in U(15)\), s.t. \(|<a>|=8\), \(U(15)\) is not a cyclic group.◻

 

 

Example \(\PageIndex{3}\)

Prove that \((U(14),(\bullet \pmod{14}))\) is cyclic. 

Proof:

We will show that \(\exists a \in U(14)\) s.t. \(|<a>|=|U(14)|\).

Consider  \(U(14)=\{1,3,5,9,11,13\}\) where \(|U(14)|=6\).

Consider \(3^1\equiv 3 \pmod{14}\), \(3^2\equiv 9 \pmod{14}\), \(3^3 \equiv 13 \pmod{14}\),  \(3^4\equiv 11 \pmod{14}\),  \(3^5\equiv 5 \pmod{14}\) and \(3^6 \equiv 1 \pmod{14}\).

\(|<3>|=6\) since \(3^6 \equiv 1 \pmod{14}\).

Thus \(|<3>|=6\).

Therefore, \(U(14)\) is cyclic and \(3\) is a generator.◻

Example \(\PageIndex{4}\)

Is \((\mathbb{Z},+)\) cyclic group?  If so, what are the possible generators?

Yes, \((\mathbb{Z},+)\) is a cyclic group that is generated by \(\pm 1\).

Proof of being a cyclic group:

Since the group generated by \(1\) contains \(1,\)  the identity  \(0,\)  and the inverse of \(1, (-1),\) as well as all multiples of \(1 \) and \((-1)\), \((\mathbb{Z},+)\) is cyclic.

 

Possible Generators:

Note \(1^n\) is \(1+1+\cdots+1\) with \(n\) terms when \(n>0\) and \((-1)+(-1)+\cdots+(-1)\) with \(n\) terms when \(n\) is \(<0\).

We shall show \(\mathbb{Z}=<1>=<-1>\).

Consider that \(1^n\) means \(1+1+\cdots +1\) with \(n\) terms and that \(1^{-n}\) means \(+(-1)+(-1)+\cdots+(-1)\) with \(n\) terms.  

It should be clear that \(1^n\) will generate \(\mathbb{Z}_+\) and that \(1^{-n}\) will generate \(\mathbb{Z}_-\).  Note that \(1^0\) is interpreted as \(0\cdot 1=0\).  

\(<1>=\{ \ldots, -3,-2,-1,0,1,2,3,\ldots \}\).

Thus \(<1>\) is a generator of \((\mathbb{Z},+)\).

Similarly, we will show that \(<-1>\) is a generator of \((\mathbb{Z},+)\).

 Consider that \((-1)^n\) means \(+(-1)+(-1)+\cdots+(-1)\) with \(n\) terms and that \(1^{-n}\) means \(1+1+\cdots +1\) with \(n\) terms.  

It should be clear that \(1^n\) will generate \(\mathbb{Z}_-\) and that \(1^{-n}\) will generate \(\mathbb{Z}_+\).  \(<-1>=\{ \ldots, -3,-2,-1,0,1,2,3,\ldots \}\).

Thus the generators of \((\mathbb{Z},+)\) are \(<1>\) and \(<-1>\).

Properties: 

Theorem \(\PageIndex{2}\)

Cyclic groups are abelian.

Proof:

Let \((G,\star)\) be a group.

If \(G\) is cyclic, then \(G\) is abelian.

Assume that \(G\) is cyclic.

Then \(G=<g>\) for some \(g \in G\).

Let \(a,b \in G\).

We will show \(ab=ba\).

Then \(a=g^m\) and \(b=g^n\), \(m,n \in \mathbb{Z}\).

Consider \(ab=g^mg^n\)

        \(=g^{m+n}\)

        \(=g^{n+m}\)

        \(=g^ng^m\)

        \(=ba\)

Hence \(G\) is abelian.◻

Note

The converse is not true.  Specifically, if a group is abelian, it is not necessarily cyclic.  A counterexample is \((U(15), \bullet)\) since it is abelian but not cyclic.

 

Theorem \(\PageIndex{3}\)

Every subgroup of a cyclic group will be cyclic.

Let \((G, \star)\) be a cyclic group.  If \(H \le G\), then \(H\) is a cyclic group.

Proof

Let \(G=<g>, \; g \in G\).

We will show \(H=<g^k>\) for some \(k \in \mathbb{Z}\).

There are two cases to consider.

Case 1:  \(H=\{e\}\)

Let \(H=\{e\}\), then we are done since \(e \star e=e\) thus \(|H|=|<e>|=1\).

Case 2:  \(H \ne \{e\}\) 

Let \(H \ne \{e\}\).

Thus \(\exists h \in H\) s.t. \(h \ne e\).

Since \(h \in H\), \(h \in G\).

Hence \(h=g^m\) for some \(m \in \mathbb{Z}\).  

Without loss of generality, we may assume \(m \in \mathbb{N}\).

Define \(S=\{n\in \mathbb{N}|g^n \in H\} \subseteq \mathbb{N}\).

Since \(S(\ne \{\})\) by the well ordering principle \(S\) has a smallest element \(k\).

 

Let \(k \in \mathbb{Z}\).

We shall show that \(H=<g^k>\).

Since \(g^k \in H\), \(<g^k> \le H\).

Let \(h \in H\).

We shall show that \(h \in <g^k>\).

Since \(h \in H\), \(h=g^m, m \in \mathbb{Z}\).

Since \(k,m \in \mathbb{Z}\), by the division algorithm, there exists \(q,r\) s.t. \(m=qk+r, \; 0\le r<k\).

Then \(g^m=g^{qk}g^r\) and \(g^{m-qk}=g^r\).

Since \(g^m, g^{qk} \in H\), then \(g^r \in H\).

Since \(k\) is the smallest, \(r=0\).  (I think this is because \(g^0=e\))

Therefore \(g^m=(g^k)^q \in <g^k>\).

Thus \(H \le <g^k>\).◻

 

Note

The Converse is not true. Every sub-group of the Klein 4-group is cyclic, but not the group.

Corollary \(\PageIndex{4}\)

Subgroups of \( (\mathbb{Z}, +)\) are of the form \((n\mathbb{Z},+)\), \(n\in \mathbb{N}\).

Theorem \(\PageIndex{5}\)

Let \(G\) be a cyclic group s.t. \(G=<a>\) and \(|G|=n\).

Then \(a^k=e\) iff \(n|k\).

Proof

Let \(G=<a>\) with \(a^n=e\).

\(<a>=\{e,a,a^2\ldots,a^{n-1}\}\).

Let \(a^k=e, \; k \in \mathbb{Z}\).

Since by the division algorithm, unique integers q and r exist such that   \(k=nq+r, \; 0\le r<n, \; q,r \in \mathbb{Z}\).

Consider \(a^r=a^{k-nq}\) 

        \(=a^k(a^n)^{-q}\) Note: \(a^k=e\) is our assumption & \(n\) is the order of group.

        \(=ee^{-q}\)

        \(=e\).

Hence \(r=0\).

Thus \(k=nq\).

Therefore \(n|k\). 

Conversely, assume that \(n|k\).

Then \(k=nm, m \in \mathbb{Z}\).

\(a^k=a^{nm}\).

     \(=(a^n)^m\)

     \(=e^m\)

     \(=e\).

Hence the result.◻

 

Theorem \(\PageIndex{6}\)

\(<a^k>=<a^{\gcd(n,k)}>\)

Let \(G=<a>\) be a cyclic group with \(|G|=n\).

Let \(b \in G\).

If \(b=a^k\) then \(|b|=\frac{n}{d}, \; d=\gcd(k,n)\).

 

Answer

Let \(G=<a>\) with \(|G|=n\) and \(a \in G\).

Let \(a^n=e\).

We will show that \(<a^k>=<a^{\gcd{(n,k)}}>\).

Let \(x \in <a^k>\).

Since \(\gcd(n,k)=d\) for some \(d \in \mathbb{Z}_+\), \(d|n\) and \(d|k\).

Thus, \( k=md, m\in \mathbb{Z}\).

Consider an arbitrary power of \(a^k\), \((a^k)^j \in <a^k>\).

\)(a^k)^j = (a^{md})^j=(a^d)^{jm} \in <a^d> \in <a^{\gcd{(n,k)}}>\).

Let \(d=xn+yk, \; n,k \in \mathbb{Z}\).

Using the division algorithm, \((a^{xn+yk})^h= (a^n)^{xh} (a^k)^{yh}, \;n,k,h \in \mathbb{Z}\)

    \(=(a^k)^{yh} \in <a^k>\).

Thus \(a^k = a^d\).

Now we want to show \(|a^k|=\frac{n}{d}\).

Consider \((a^k)^{\frac{n}{d}}=(a^n)^{\frac{k}{d}}=e^{\frac{k}{d}}=e\).

Thus \(|a^k|\le \frac{n}{d}\).

Now consider \((a^d)^{\frac{n}{d}}=(a^n)=e \; \Rightarrow \; |a^d| \le \frac{n}{d}\).

Now consider \(0 \le i <\frac{n}{d}\).

Thus \( di < n\), where \(d\) is positive since it is the greatest common denominator.

Thus \((a^d)^i=a^{di}\ne e\) for \(di<n\).

Then we can show \(|a^d|=\frac{n}{d}\).

Consider \(|a^k|=|<a^k>|=|<a^d>|=|a^d|= \frac{n}{d}\).

Hence \(|a^k|=\frac{n}{d}\).

 

Theorem \(\PageIndex{7}\)

A group of prime order is cyclic.

Proof:

Let \(G\) be a group s.t. \(|G|=p\), where \(p\) is a prime number.
Let \(g \in G\) s.t. \(<g>\) is a subgroup of \(G\).

Since \(p>1\), \(g \ne \{e\}\).

If \(G\) is cyclic,  \(\exists \; g^k=e\), for some \( k \in \mathbb{Z}\).  

By the division algorithm, \(k=nq+r\) where \(0 \le r <n\).

Hence, \(e=g^k=g^{nq+r}=g^{nq}g^r=eg^r=g^r\).

Since the smallest positive integer \(k\) such that \(g^k=e\) is \(n\), \(r=0\).  Thus \(n|k\).

Conversely, if \(n|k\), then \(k=nm\), \(m \in \mathbb{Z}\). 

Consequently, \(g^k=g^{nm}=(g^n)^m=e^m=e\).

Thus, \(|<g>|\) divides \(|G|\) if \(G\) is cyclic.

Consider that the only two divisors of a prime number are the prime number itself and 1.

Since \(p > 1\), \(|G|=p\).

Since \(|G|=p\), \(G\) is cyclic.◻

 

 

Example \(\PageIndex{6}\)

1. List the cyclic subgroups of \(U(30)\).

\(U(30)=\{1,7,11,13,17,19,23,29\}\).

Thus \(|U(30)|=8\).

\(k\)	Calculations	\(|k|\)
7	\(7^1 \equiv 7 \pmod{30}\), \(7^2 \equiv 19 \pmod{30}\), \(7^3 \equiv 13 \pmod{30}\), \(7^4 \equiv 1 \pmod{30}\)	4
11	\(11^1 \equiv 11 \pmod{30}\), \(11^2 \equiv 1 \pmod{30}\)	2
13	\(13^1 \equiv 13 \pmod{30}\), \(13^2 \equiv 19 \pmod{30}\), \(13^3 \equiv 7 \pmod{30}\), \(13^4 \equiv 1 \pmod{30}\)	4
17	\(17^1 \equiv 17 \pmod{30}\), \(17^2 \equiv 19 \pmod{30}\), \(17^3 \equiv 23 \pmod{30}\), \(17^4 \equiv 1 \pmod{30}\)	4
19	\(19^1 \equiv 19 \pmod{30}\), \(19^2 \equiv 1 \pmod{30}\)	2
23	\(23^1 \equiv 23 \pmod{30}\), \(23^2 \equiv 19 \pmod{30}\), \(23^3 \equiv 17 \pmod{30}\),  \(23^4 \equiv 1 \pmod{30}\)	4
29	\(29^1 \equiv 29 \pmod{30}\), \(29^2 \equiv 1 \pmod{30}\)	2

 

Since \(\not \exists a \in U(30)\) s.t. \(|<a>|=|U(30)|=8\), \(U(30)\) is not a cyclic group.

However the following subgroups of \(U(30)\) are cyclic: \(\{1,7,13,19\}\), \(\{1,17,19,23\}\), \(\{1,11\}\),  \( \{1,19\} \), \(\{1,29\}\) and \(\{1\}\).