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.

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 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~\(a\text{.}\)

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 \(a\text{.}\)

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 \({\mathbb Z}_6\text{.}\)

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 Figure 4.1.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.

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,
\end{equation*}