Cyclic Groups

Definition. A group $G$ is cyclic if there is an element $g \in G$ such that all elements of $G$ are powers of $g$. Such an element is called a generator.

\[G = \set{g^n : n \in \mathbb{Z}}\]

Cyclic Subgroup

Definition. For a group $G$ and $a \in G$, the cyclic subgroup generated by $a$, denoted by $\langle a \rangle$, is defined to be the smallest subgroup containing $a$.

Proposition. $\langle a \rangle$ is the set of all powers of $a$.

Proof.

For the set of all powers of $a$, $a^0 = e$ is a power of $a$, so it is non-empty. Given $a^m$ and $a^n$, $(a^m)(a^n)^{-1} = a^{m-n}$ is a power of $a$. So, the powers of $a$ is a subgroup of $G$.

If $a$ is in any subgroup, the inverse $a^{-1}$ has to be in the subgroup, and hence $a^n$ for all integers $n$ has to be in the subgroup as well by closure. As the powers of $a$ is already a subgroup from the above, it is in fact the smallest one containing $a$.

Proposition. For $a \in G$, $\text{ord}(a) = | \langle a\rangle |$.

Proof.

If $\text{ord}(a) = \infty$, then $a^n \not= e$ for any $n$ and $a^m \not = a^n$ for all $m \not = n$, hence $\langle a \rangle$ has infinite order.

If $\text{ord}(a) = n$, then $\langle a \rangle = \set{a^0 = e, a, a^1, …, a^{n-1}}$, as $a^n = e$, $a^{n+1} = a$, and so on, also $a^{-1} = a^{n-1}$ etc. So $|\langle a \rangle| = n$.

Isomorphism

Theorem. Any two cyclic groups of the same finite order $n$ are isomorphic, and hence isomorphic to $(\mathbb{Z}_n, +)$.

Proof.

Suppose $G$ is cyclic group of order $n$ and $g \in G$ is the generator. Consider the map $f: \mathbb{Z}_n \to G$ defined by $f(m) = g^m$.

\[f(a + b) = g^{a+b} = (g^a)(g^b) = f(a)f(b)\]

so $f$ is a homomorphism. As $G = \set{g^0, g^1, g^2, …, g^{n-1}}$, $f$ is surjective. As $|G| = |\mathbb{Z}_n| = n$, surjectivity implies injectivity. Hence, $f$ is a isomorphism and $G$ is isomorphic to $\mathbb{Z}_n$, and therefore to each other.

Theorem. Any two infinite cyclic groups are isomorphic, and hence isomorphic to $(\mathbb{Z}, +)$

Proof.

Let $G$ be an infinite cyclic group generated by $g \in G$. Consider the map $f: \mathbb{Z} \to G$ defined by $f(m) = g^m$. Similarily,

\[f(a + b) = g^{a+b} = (g^a)(g^b) = f(a)f(b)\]

so $f$ is a homomorphism. $f$ is surjective by definition. If $f(a) = f(b)$, then $g^a = g^b$, which implies $a = b$, otherwise $g^{a-b}=e$ and $G$ would be finite. Hence, $f$ is a isomorphism and $G$ is isomorphic to $(\mathbb{Z}, +)$.

So to conclude, there is only one cyclic group of order $n$, and is usually denoted by $C_n$.

Alternatively, we can combine the above by Isomorphism Theorem.

Theorem. Any cyclic group is isomorphic to either $(\mathbb{Z}, +)$ or $(\mathbb{Z}/n\mathbb{Z}, +)$ for some $n \in \mathbb{N}$.

Proof.

Let $G = \langle g \rangle$ be cyclic and $f: \mathbb{Z} \to G$ defined by $f(m) = g^{m}$. $f$ is a surjective homomorphism.

Consider the $\ker f \trianglelefteq \mathbb{Z}$, either

  • $\ker f = \set{e}$, so $f$ is isomorphism and $G \cong \mathbb{Z}$.

  • $\ker f = \mathbb{Z}$, so $G \cong \mathbb{Z}/\mathbb{Z} = \set{e} = C_1$.

  • $\ker f = n\mathbb{Z}$ for some $n \in \mathbb{N}$, so $G \cong \mathbb{Z}/n\mathbb{Z} \cong C_n$.

Properties

Property. All cyclic groups are abelian.

Proof.

Let $G$ be a cyclic group and $g \in G$ generates $G$. For any two elements $a = g^m$ and $b = g^n$, we have

\[ab = (g^m)(g^n) = g^{m+n} = g^{n+m} = (g^n)(g^m) = ba\]

Hence, $G$ is abelian.

In short, as $C_n \cong \mathbb{Z}_n$, $C_n$ shares the commutativity with additive group $\mathbb{Z}_n$.

Property. A subgroup of cyclic group is cyclic.

Proof.

Let $G$ be a cyclic group and $H \le G$ and $g \in G$ be the generator. Therefore, $g^n \in H$ for some $n$. Let $d$ be the smallest positive integer such that $g^d \in H$, by closure $g^{md} \in H$ for $m \in \mathbb{Z}$. For an element $g^{md + r} \in H$, where $0 \le r < d$. As $g^{-md} \in H$, $g^r \in H$, which is only possible when $r = 0$. Hence, $H = \set{g^{md} : m \in \mathbb{Z}}$ is cyclic.

Property. A group of prime order is cyclic, and is generated by any of its elements other than the identity element.

Proof.

Let $G$ be a group with order $p$ where $p$ is prime. For any $g \in G$, by Lagrange Theorem, $\text{ord}(g) \mid p$. If $\text{ord}(g) = 1$, $g = e$. Otherwise, $\text{ord}(g) = p$ and generates $G$. Hence, $G$ is cyclic and is generated by any elements other than $e$.

Property. $C_m \times C_n \cong C_{mn}$ iff $m$ and $n$ are relatively prime.

Proof.

Let $C_m$ be generated by $a$ and $C_n$ is generated by $b$. We have $(a, b)^k = (a^k, b^k) = e$ iff $a^k = e$ and $b^k = e$. If $m$ and $n$ are relatively prime, it happens for the first time when $k = mn$. Hence, the order of $(a, b)$ is $mn$ and generates $C_{mn}$

If $(m, n) = d$, for $k = mn/d$, $(a, b)^k = (e, e) = e$ for all element $(a, b)$. So, there is no element of order $mn$ and $C_m \times C_n \not \cong C_{mn}$.

Examples

$(\mathbb{Z}, +)$ is not cyclic. However,

Proposition. A subgroup of $(\mathbb{Z}, +)$ is cyclic iff it has a smallest positive element $h$, in which is the generator.

Proof.

$(\Rightarrow)$ Let $H$ be a cyclic group, say $H = \set{nh : n \in \mathbb{Z}}$, then $|h|$ is the smallest positive element.

$(\Leftarrow)$ If $H$ has a smallest positive element $h$, then by closure, $nh \in H$ for $n \in \mathbb{Z}$. For an element $a = nh + r \in H$, where $0 \le r < h$. As $-nh \in H$, we have $a - nh = r \in H$, but $h$ is the smmalest positive element in $H$ so $r = 0$. Hence, $H = \set{nh : n \in \mathbb{Z}}$.

References