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Section4.1Cyclic Subgroups

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.

Example4.1

Suppose that we consider \(3 \in {\mathbb Z}\) and look at all multiples (both positive and negative) of 3. As a set, this is \begin{equation*}3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}.\end{equation*} 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.

Example4.2

If \(H = \{ 2^n : n \in {\mathbb Z} \}\), then \(H\) is a subgroup of the multiplicative group of nonzero rational numbers, \({\mathbb Q}^*\). If \(a = 2^m\) and \(b = 2^n\) are in \(H\), then \(ab^{-1} = 2^m 2^{-n} = 2^{m-n}\) is also in \(H\). By Proposition 3.31, \(H\) is a subgroup of \({\mathbb Q}^*\) determined by the element 2.

Proof
Remark4.4

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

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

Example4.5

Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate \({\mathbb Z}_6\); 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 \(\langle 2 \rangle = \{ 0, 2, 4 \}\).

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

Example4.6

The group of units, \(U(9)\), in \({\mathbb Z}_9\) is a cyclic group. As a set, \(U(9)\) is \(\{ 1, 2, 4, 5, 7, 8 \}\). The element 2 is a generator for \(U(9)\) since \begin{align*} 2^1 & = 2 \qquad 2^2 = 4\\ 2^3 & = 8 \qquad 2^4 = 7\\ 2^5 & = 5 \qquad 2^6 = 1. \end{align*}

Example4.7

Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle \(S_3\). The multiplication table for this group is Table 3.7. The subgroups of \(S_3\) are shown in Figure 4.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.

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Figure4.8Subgroups of \(S_3\)
Proof

SubsectionSubgroups of Cyclic Groups

We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If \(G\) is a group, which subgroups of \(G\) are cyclic? If \(G\) is a cyclic group, what type of subgroups does \(G\) possess?

Proof
Proof
Example4.15

Let us examine the group \({\mathbb Z}_{16}\). The numbers 1, 3, 5, 7, 9, 11, 13, and 15 are the elements of \({\mathbb Z}_{16}\) that are relatively prime to 16. Each of these elements generates \({\mathbb Z}_{16}\). For example, \begin{align*} 1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11\\ 4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6 \\ 7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1 \\ 10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12\\ 13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7. \end{align*}