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.

Theorem4.3

Let \(G\) be a group and \(a\) be any element in \(G\). Then the set
\begin{equation*}\langle a \rangle = \{ a^k : k \in {\mathbb Z} \}\end{equation*}
is a subgroup of \(G\). Furthermore, \(\langle a \rangle\) is the smallest subgroup of \(G\) that contains \(a\).

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

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.

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

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?

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let \(G\) be a cyclic group generated by \(a\) and suppose that \(H\) is a subgroup of \(G\). If \(H = \{ e \}\), then trivially \(H\) is cyclic. Suppose that \(H\) contains some other element \(g\) distinct from the identity. Then \(g\) can be written as \(a^n\) for some integer \(n\). Since \(H\) is a subgroup, \(g^{-1} = a^{-n}\) must also be in \(H\). Since either \(n\) or \(-n\) is positive, we can assume that \(H\) contains posiitve powers of \(a\) and \(n \gt 0\). Let \(m\) be the smallest natural number such that \(a^m \in H\). Such an \(m\) exists by the Principle of Well-Ordering.

We claim that \(h = a^m\) is a generator for \(H\). We must show that every \(h' \in H\) can be written as a power of \(h\). Since \(h' \in H\) and \(H\) is a subgroup of \(G\), \(h' = a^k\) for some integer \(k\). Using the division algorithm, we can find numbers \(q\) and \(r\) such that \(k = mq +r\) where \(0 \leq r \lt m\); hence,
\begin{equation*}a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r.\end{equation*}
So \(a^r = a^k h^{-q}\). Since \(a^k\) and \(h^{-q}\) are in \(H\), \(a^r\) must also be in \(H\). However, \(m\) was the smallest positive number such that \(a^m\) was in \(H\); consequently, \(r=0\) and so \(k=mq\). Therefore,
\begin{equation*}h' = a^k = a^{mq} = h^q\end{equation*}
and \(H\) is generated by \(h\).

Corollary4.11

The subgroups of \({\mathbb Z}\) are exactly \(n{\mathbb Z}\) for \(n = 0, 1, 2,\ldots\).

Proposition4.12

Let \(G\) be a cyclic group of order \(n\) and suppose that \(a\) is a generator for \(G\). Then \(a^k=e\) if and only if \(n\) divides \(k\).

First suppose that \(a^k=e\). By the division algorithm, \(k = nq + r\) where \(0 \leq r \lt n\); hence,
\begin{equation*}e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r.\end{equation*}
Since the smallest positive integer \(m\) such that \(a^m = e\) is \(n\), \(r= 0\).

Conversely, if \(n\) divides \(k\), then \(k=ns\) for some integer \(s\). Consequently,
\begin{equation*}a^k = a^{ns} = (a^n)^s = e^s = e.\end{equation*}

Theorem4.13

Let \(G\) be a cyclic group of order \(n\) and suppose that \(a \in G\) is a
generator of the group. If \(b = a^k\), then the order of \(b\) is \(n/d\), where \(d = \gcd(k,n)\).

We wish to find the smallest integer \(m\) such that \(e = b^m = a^{km}\). By Proposition 4.12, this is the smallest integer \(m\) such that \(n\) divides \(km\) or, equivalently, \(n/d\) divides \(m(k/d)\). Since \(d\) is the greatest common divisor of \(n\) and \(k\), \(n/d\) and \(k/d\) are relatively prime. Hence, for \(n/d\) to divide \(m(k/d)\) it must divide \(m\). The smallest such \(m\) is \(n/d\).

Corollary4.14

The generators of \({\mathbb Z}_n\) are the integers \(r\) such that \(1 \leq r \lt n\) and \(\gcd(r,n) = 1\).

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*}