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Section 11.1 Cyclic Groups and Orders of Elements

Given an element \(a\) in a group \(G\text{,}\) we can consider what happens when we multiply \(a\) by itself (or if we are writing the group additively, add \(a\) to itself). Depending on what group we are in, if we continue to multiply the result by \(a\text{,}\) we might eventually find ourselves back at the identity. How long does this take? That is precisely the question that the order of a group element addresses. We will see that this relatively simple idea has far reaching consequences.

Worksheet 11.1.1 Powers mod Powers

Main Question: What is the remainder when you divide \(a^p\) by \(p\text{?}\)


Compute \(a^p\) and its remainder when divided by \(p\text{,}\) for various values of \(a\) and \(p\text{.}\) Everyone should do at least 5, and then share with the group. As a group, discuss any patterns you see and form a conjecture.

Find the remainders when you perform the following divisions. Try different values of \(a\text{.}\) You should first guess what the value is based on your conjecture and then verify (or refute) your guess.


\(a^6\) divided by 10?


\(a^9\) divided by 15?


\(a^{13}\) divided by 21?


\(a^{2321}\) divided by \(2419\text{?}\)

Discuss in your groups: how might we think about the main question here in terms of group theory? What would we need to prove (about groups)?

Worksheet 11.1.2 Activity: Orders of Permutations

The order of an element \(g\) in a group \(G\) is the least natural number \(n\) such that \(g^n = e\text{,}\) if such a number exists (otherwise we say the order of \(g\) is infinite).


Find the orders of the elements of \(S_5\) below:

\begin{equation*} \alpha = (12) \hspace{1in}\alpha = (123) \hspace{1in} \alpha = (1234) \hspace{1in} \alpha = (12345) \end{equation*}

The orders are 2, 3, 4, and 5, respectively.


Find an element of \(S_5\) that has an order different from those found above.


We will need an element that is not just a cycle, since the longest cycle is of length 5. So how about \((12)(345)\text{.}\) This will have order 6.


Let \(\alpha\) be an element of \(S_5\text{.}\) What is \(\alpha^{120}\text{?}\)


No matter what element we pick, \(\alpha^{120}\) will always be \((1)\text{.}\) This will work in general, but you can also see this by considering the orders of elements, which must all divide the size of the group.


Is there an element in \(S_5\) that has order \(120\text{?}\)


No there is not. If there were, this would say that \(S_5\) was cyclic, and would thus be abelian, which we know is not the case. Alternatively, just look at how to write elements as disjoint cycles.


What is the largest order of any element in \(S_5\text{?}\)


The largest order will be 6, for elements that are the disjoint product of a 2-cycle and 3-cycle.

Subsection 11.1.3 Cyclic 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.

Example 11.1.

Suppose that we consider \(3 \in {\mathbb Z}\) and look at all multiples (both positive and negative) of \(3\text{.}\) As a set, this is

\begin{equation*} 3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}\text{.} \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 \(3\text{.}\) Every element in the subgroup is “generated” by \(3\text{.}\)

Example 11.2.

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 \(H\text{.}\) Thus \(H\) is a subgroup of \({\mathbb Q}^*\) determined by the element \(2\text{.}\)


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 \(a\text{.}\)

Remark 11.4.

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 \(a\text{.}\)

Example 11.5.

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 \(3\text{.}\) 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 \({\mathbb Z}_6\text{.}\)

Example 11.6.

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

\begin{align*} 2^1 & = 2 \qquad 2^2 = 4\\ 2^3 & = 8 \qquad 2^4 = 7\\ 2^5 & = 5 \qquad 2^6 = 1\text{.} \end{align*}
Example 11.7.

Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle \(S_3\text{.}\) The multiplication table for this group is Figure 2.7. The subgroups of \(S_3\) are shown in Figure 2.39. Notice that every subgroup is cyclic; however, no single element generates the entire group.

Figure 11.8. Subgroups of \(S_3\)

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*}

\(G\) is abelian.

Subsection 11.1.4 Subgroups 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?


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 \(G\text{.}\) If \(H = \{ e \}\text{,}\) 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 \(n\text{.}\) Since \(H\) is a subgroup, \(g^{-1} = a^{-n}\) must also be in \(H\text{.}\) Since either \(n\) or \(-n\) is positive, we can assume that \(H\) contains positive powers of \(a\) and \(n \gt 0\text{.}\) Let \(m\) be the smallest natural number such that \(a^m \in H\text{.}\) Such an \(m\) exists by the Principle of Well-Ordering.

We claim that \(h = a^m\) is a generator for \(H\text{.}\) We must show that every \(h' \in H\) can be written as a power of \(h\text{.}\) Since \(h' \in H\) and \(H\) is a subgroup of \(G\text{,}\) \(h' = a^k\) for some integer \(k\text{.}\) Using the division algorithm, we can find numbers \(q\) and \(r\) such that \(k = mq +r\) where \(0 \leq r \lt m\text{;}\) hence,

\begin{equation*} a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r\text{.} \end{equation*}

So \(a^r = a^k h^{-q}\text{.}\) Since \(a^k\) and \(h^{-q}\) are in \(H\text{,}\) \(a^r\) must also be in \(H\text{.}\) However, \(m\) was the smallest positive number such that \(a^m\) was in \(H\text{;}\) consequently, \(r=0\) and so \(k=mq\text{.}\) Therefore,

\begin{equation*} h' = a^k = a^{mq} = h^q \end{equation*}

and \(H\) is generated by \(h\text{.}\)


First suppose that \(a^k=e\text{.}\) By the division algorithm, \(k = nq + r\) where \(0 \leq r \lt n\text{;}\) hence,

\begin{equation*} e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r\text{.} \end{equation*}

Since the smallest positive integer \(m\) such that \(a^m = e\) is \(n\text{,}\) \(r= 0\text{.}\)

Conversely, if \(n\) divides \(k\text{,}\) then \(k=ns\) for some integer \(s\text{.}\) Consequently,

\begin{equation*} a^k = a^{ns} = (a^n)^s = e^s = e\text{.} \end{equation*}

We wish to find the smallest integer \(m\) such that \(e = b^m = a^{km}\text{.}\) By Proposition 2.43, this is the smallest integer \(m\) such that \(n\) divides \(km\) or, equivalently, \(n/d\) divides \(m(k/d)\text{.}\) Since \(d\) is the greatest common divisor of \(n\) and \(k\text{,}\) \(n/d\) and \(k/d\) are relatively prime. Hence, for \(n/d\) to divide \(m(k/d)\) it must divide \(m\text{.}\) The smallest such \(m\) is \(n/d\text{.}\)

Example 11.15.

Let us examine the group \({\mathbb Z}_{16}\text{.}\) The numbers \(1\text{,}\) \(3\text{,}\) \(5\text{,}\) \(7\text{,}\) \(9\text{,}\) \(11\text{,}\) \(13\text{,}\) and \(15\) are the elements of \({\mathbb Z}_{16}\) that are relatively prime to \(16\text{.}\) Each of these elements generates \({\mathbb Z}_{16}\text{.}\) 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\text{.} \end{align*}

Exercises 11.1.5 Collected Homework


Prove the following basic facts about orders of elements. None of these are particularly difficult, so you should focus on writing nice, clean proofs. In each of the following, \(a\) is an element of a group \(G\text{.}\)


If \(\ord(a) = n\text{,}\) then for any \(r \lt n\text{,}\) the inverse of \(a^r\) is \(a^{n-r}\text{.}\) That is, \((a^r)\inv = a^{n-r}\text{.}\)


The order of \(a\inv\) is the same as the order of \(a\text{.}\)


Careful: it is not enough to say that \((a\inv)^n = e\) (where \(\ord(a) = n\)). That is only half of it, since you must also show that no smaller power of \(a\inv\) is the identity.


If \(a^k = e\) where \(k\) is odd, then the order of \(a\) is odd.


If \(a \ne e\) and \(a^p = e\) where \(p\) is prime, then \(\ord(a) = p\text{.}\)