Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.

It is easy to see that \(3 {\mathbb Z}\) is a subgroup of the integers. This subgroup is completely determined by the element 3 since we can obtain all of the other elements of the group by taking multiples of 3. Every element in the subgroup is “generated” by 3.

Example4.1.2.An Infinite Cyclic Subgroup, Multiplication of Rational Numbers.

If \(H = \{ 2^n : n \in {\mathbb Z} \}\text{,}\) then \(H\) is a subgroup of the multiplicative group of nonzero rational numbers, \({\mathbb Q}^*\text{.}\) If \(a = 2^m\) and \(b = 2^n\) are in \(H\text{,}\) then \(ab^{-1} = 2^m 2^{-n} = 2^{m-n}\) is also in \(H\text{.}\) By Proposition 3.3.8, \(H\) is a subgroup of \({\mathbb Q}^*\) determined by the element 2.

Theorem4.1.3.

Let \(G\) be a group and \(a\) be any element in \(G\text{.}\) Then the set

\begin{equation*}
\langle a \rangle = \{ a^k : k \in {\mathbb Z} \}
\end{equation*}

is a subgroup of \(G\text{.}\) Furthermore, \(\langle a \rangle\) is the smallest subgroup of \(G\) that contains~\(a\text{.}\)

Proof.

The identity is in \(\langle a \rangle \) since \(a^0 = e\text{.}\) If \(g\) and \(h\) are any two elements in \(\langle a \rangle \text{,}\) then by the definition of \(\langle a \rangle\) we can write \(g = a^m\) and \(h = a^n\) for some integers \(m\) and \(n\text{.}\) So \(gh = a^m a^n = a^{m+n}\) is again in \(\langle a \rangle \text{.}\) Finally, if \(g = a^n\) in \(\langle a \rangle \text{,}\) then the inverse \(g^{-1} = a^{-n}\) is also in \(\langle a \rangle \text{.}\) Clearly, any subgroup \(H\) of \(G\) containing \(a\) must contain all the powers of \(a\) by closure; hence, \(H\) contains \(\langle a \rangle \text{.}\) Therefore, \(\langle a \rangle \) is the smallest subgroup of \(G\) containing \(a\text{.}\)

Remark4.1.4.

If we are using the “+” notation, as in the case of the integers under addition, we write \(\langle a \rangle = \{ na : n \in {\mathbb Z} \}\text{.}\)

For \(a \in G\text{,}\) we call \(\langle a \rangle \) the cyclic subgroup generated by \(a\text{.}\) If \(G\) contains some element \(a\) such that \(G = \langle a \rangle \text{,}\) then \(G\) is a cyclic group. In this case \(a\) is a generator of \(G\text{.}\) If \(a\) is an element of a group \(G\text{,}\) we define the order of \(a\) to be the smallest positive integer \(n\) such that \(a^n= e\text{,}\) and we write \(|a| = n\text{.}\) If there is no such integer \(n\text{,}\) we say that the order of \(a\) is infinite and write \(|a| = \infty\) to denote the order of \(a\text{.}\)

Example4.1.5.Generators of a Finite Cyclic Group.

Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate \({\mathbb Z}_6\text{;}\) hence, \({\mathbb Z}_6\) is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of \(2 \in {\mathbb Z}_6\) is 3. The cyclic subgroup generated by 2 is \(\langle 2 \rangle = \{ 0, 2, 4 \}\text{.}\)

The groups \({\mathbb Z}\) and \({\mathbb Z}_n\) are cyclic groups. The elements 1 and \(-1\) are generators for \({\mathbb Z}\text{.}\) We can certainly generate \({\mathbb Z}_n\) with 1 although there may be other generators of \({\mathbb Z}_n\text{,}\) as in the case of \({\mathbb Z}_6\text{.}\)

Example4.1.6.A Cyclic Group of Units.

The group of units, \(U(9)\text{,}\) in \({\mathbb Z}_9\) is a cyclic group. As a set, \(U(9)\) is \(\{ 1, 2, 4, 5, 7, 8 \}\text{.}\) The element 2 is a generator for \(U(9)\) since

Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle \(S_3\text{.}\) The subgroups of \(S_3\) are shown in Figure 4.1.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.

Theorem4.1.9.

Every cyclic group is abelian.

Proof.

Let \(G\) be a cyclic group and \(a \in G\) be a generator for \(G\text{.}\) If \(g\) and \(h\) are in \(G\text{,}\) then they can be written as powers of \(a\text{,}\) say \(g = a^r\) and \(h = a^s\text{.}\) Since

\begin{equation*}
g h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g\text{,}
\end{equation*}