Introduction to Cyclic Groups

Definition: order of an element

Let \((G,⋆)\) be a group.

Let \(a ∈ G\), then the order of the element \(a ∈ G\), 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 \((G,⋆)\) be a group.

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

Theorem 2.4.1

Let \((G,⋆)\) be a group.

Let \(a∈G\), then

\(<a>≤G\).

Example 2.4.1

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

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

Note

Let \((G,⋆)\) be a group.

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

Definition: Cyclic subgroup

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

Example 2.4.1

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

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

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

We will show that \(∄a∈U(15)\), s.t. \(|<a>|=8\).

\(|<1>|=1\) since \(1^1≡1 \pmod{15}\)

\(|<2>|=4\) since \(2^4≡1 \pmod{15}\)

\(|<4>|=2\) since \(4^2≡1 \pmod{15}\)

\(|<7>|=4\) since \(7^4≡1 \pmod{15}\)

\(|<8>|=4\) since \(8^4≡1 \pmod{15}\)

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

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

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

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

Example 2.4.2

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

Proof:

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

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

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

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

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

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

Example 2.4.1

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

Yes, (\mathbb{Z},+) is a cyclic group that is generated by ±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+⋯+1 with n terms when \(n > 0\) and (−1)+(−1)+⋯+(−1) with n terms when n is <0.

We shall show \(Z=<1>=<−1>\).

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

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

\(<1>={…,−3,−2,−1,0,1,2,3,…}\).

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

Similarly, we will show that <−1>is a generator of \((Z,+)\).

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

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

Thus the generators of \((Z,+)\) are <1> and <−1>.

Properties:

Theorem 2.4.2

Cyclic groups are abelian.

Proof:

Let \((G,⋆)\) be a group.

If G is cyclic, then G is abelian.

Assume that G is cyclic.

Then \(G=<g>\) for some \(g ∈ G\).

Let \(a, b ∈ G\).

We will show \(ab=ba\).

Then \(a=g^m\) and \(b=g^n, m, n ∈ Z\).

Consider \(ab=g^m g^n\)
        \(=g^{m+n}\)
        \(=g^{n+m}\)
        \(=g^ng^m\)
        \(=ba\)

Hence G is abelian.

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

Theorem 2.4.3

A subgroup of a cyclic group will be cyclic.

Let \((G,⋆)\) be a cyclic group.  If \(H≤G\), then H is a cyclic group.

Proof

Let \(G=<g>,g ∈ G\).

We will show \(H=<g^k>\) for some \(k ∈ Z\).

There are two cases to consider.

Case 1:  \(H={e}\)

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

Case 2:  \(H≠{e}\)

Let \(H≠{e}\).

Thus \(∃h ∈ H\) s.t. \(h≠e\).

Since \(h ∈ H, h ∈ G\).

Hence \(h=g^m\) for some \(m ∈ Z\).  

Without loss of generality, we may assume \(m ∈ N\).

Define \(S={n ∈ N | g^n ∈ H} ⊆ N\).

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

Let \(k ∈ Z\).

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

Since \(g^k ∈ H, < g^k >≤H\).

Let \(h∈H\).

We shall show that \(h ∈ <g^k>\).

Since \(h∈H\), \(h=g^m\),\(m ∈ Z\).

Since \(k, m ∈ Z \), by the division algorithm, there exists \(q ,r\) s.t. \(m=qk+r\), \(0 ≤ r < k\).

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

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

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

Therefore \(g^m=(g^k)^q ∈ <g^k>\).

Thus \(H ≤ <g^k>\).

Theorem 2.4.4

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

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

Proof

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

\(<a>={e,a,a^2…,a^{n−1}}\).

Let \(a^k = e, k ∈ Z\).

Since by the division algorithm, unique integers q and r exist such that   \(k=nq+r\), \(0 ≤ r < n, q, r ∈ Z\).

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 \(k = nm, m ∈ Z\).

\(a^k=a^{nm}\).
     \(=(a^n)^m\)
     \(=e^m\)
     \(=e\).

Hence the result.

Theorem 2.4.5

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

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

Let \(b ∈ G\).

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

Answer

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

Let \(a^n=e\).

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

Let \(x∈<ak>\).

Since \(gcd(n,k)=d\) for some \(d∈Z+\), \(d|n\) and \(d|k\).

Thus, \(k=md\), \(m ∈ Z\).

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

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

Let \(d=xn+yk,n,k ∈ Z\).

Using the division algorithm, \((a^{xn+yk})^h=(a^n)^{xh}(a^k)^{yh},n ,k ,h ∈ Z  =(a^k)^{yh} ∈ <a^k>\).

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

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

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

Thus \(|a^k| ≤ \dfrac{n}{d}\).

Now consider \((a^d)\dfrac{n}{d}=(a^n)=e ⇒ |a^d| ≤ \dfrac{n}{d}\).

Now consider \(0 ≤ i < \dfrac{n}{d}\).

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

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

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

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

Hence \(|ak|=\dfrac{n}{d}\).

Theorem 2.4.1

A group of prime order is cyclic.

Proof:

Let G be a group s.t. \(|G|=p\), where p is a prime number.

Let \(g ∈ G\) s.t. \(<g>\) is a subgroup of G.

Since \(p>1\), \(g≠{e}\).

If G is cyclic,  \(∃g^k=e\), for some \(k ∈ Z\).

By the division algorithm, \(k=nq+r\) where 0 ≤ 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 \(k=nm, m ∈ Z\). 

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 \(|G|=p\), Gis cyclic.

Example 2.4.1

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

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

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

Since \(∄a∈U(30)\) 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}.

Source: Pamini Thangarajah, https://math.libretexts.org/Courses/Mount_Royal_University/MATH_2101_Abstract_Algebra_I/Chapter_2%3A_Groups/2.4%3A_Introduction_to_cyclic_groups
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