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 $={a^m|m∈Z}$.

Theorem 2.4.1

Let $(G,⋆)$ be a group.

Let $a∈G$, then

$≤G$.

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  $G=$ 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. $||=8$.

$||=1$ since $1^1≡1 \pmod{15}$

$||=4$ since $2^4≡1 \pmod{15}$

$||=2$ since $4^2≡1 \pmod{15}$

$||=4$ since $7^4≡1 \pmod{15}$

$||=4$ since $8^4≡1 \pmod{15}$

$||=2$ since $11^2≡1 \pmod{15}$

$||=4$ since $13^4≡1 \pmod{15}$

$||=2$ since $14^2≡1 \pmod{15}$

Since $∄a∈U(15)$, s.t. $||=8$, $U(15)$ 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. $||=|U(14)|$.

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 $3^6≡1 \pmod{14}$.

$||=6$ since $3^6≡1 \pmod{14}$.

Thus $||=6$.

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 $Z==$.

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 $0 ⋅ 1=0$.  

$={…,−3,−2,−1,0,1,2,3,…}$.

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 $Z+$.  $={…,−3,−2,−1,0,1,2,3,…}$.

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 $G=$ 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 $G=,g ∈ G$.

We will show $H=$ 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 $|H|=||=1$.

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 $H=$.

Since $g^k ∈ H, < g^k >≤H$.

Let $h∈H$.

We shall show that $h ∈$.

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 $g^m=(g^k)^q ∈$.

Thus $H ≤$.

Theorem 2.4.4

Let G be a cyclic group s.t. $G=$ and $|G|=n$.

Then $a^k = e$ if $n|k$.

Proof

Let $G=$ with $a^n=e$.

$={e,a,a^2…,a^{n−1}}$.

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

$=$

Let $G=$ 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 $G=$ with $|G|=n$ and $a ∈ G$.

Let $a^n=e$.

We will show that $=$.

Let $x∈$.

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

Thus, $k=md$, $m ∈ Z$.

Consider an arbitrary power of $a^k, (a^k)^j ∈$.

$(a^k)^j = (a^{md})^j=(a^d)^{jm} \in \in$.

Let $d=xn+yk,n,k ∈ Z$.

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

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 $di$, where d is positive since it is the greatest common denominator.

Thus $(a^d)^i=a^{di} ≠ e$ for $di$.

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

Consider $|a^k| = || = || = |a^d| =\dfrac{n}{d}$.

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. 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, $||$ 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. $||=|U(30)|=8$, $U(30)$ 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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