## A Book of Abstract Algebra, Second Edition (1982)

### Chapter 10. ORDER OF GROUP ELEMENTS

Let *G* be an arbitrary group, with its operation denoted multiplicadvely. *Exponential notation* is a convenient shorthand: for any positive integer *n*, we will agree to let

and

*a*^{0} = *e*

Take care to observe that we are considering only integer exponents, *not* rational or real exponents. Raising *a* to a positive power means multiplying *a* by itself the given number of times. Raising *a* to a negative power means multiplying *a*^{−}^{1} *by* itself the given number of times. Raising *a* to the power 0 yields the group’s identity element.

These are the same conventions used in elementary algebra, and they lead to the same familiar “laws of exponents.”

**Theorem 1: Laws of exponents** *if G is a group and a ∈G*, *the following identities hold for all integers m and n:*

(i) *a ^{m}a^{n}* =

*a*

^{m}^{+n}

(ii) (*a ^{m}*)

*=*

^{n}*a*

^{mn}(iii) *a*^{−}* ^{n}* = (

*a*

^{−}

^{l})

*= (*

^{n}*a*)

^{n}^{−}

^{l}

These laws are very easy to prove when *m* and *n* are positive integers. Indeed,

(i)

(ii)

Next, by definition *a ^{−}^{n}* =

*a*

^{−}

^{1}···

*a*

^{−}

^{1}= (

*a*

^{−}

^{1})

*. Finally, since the inverse of a product is the product of the inverses in reverse order,*

^{n}(*a ^{n}*)

^{−}

^{1}= (

*aa*···

*a*)

^{−}

^{1}=

*a*

^{−}

^{1}

*a*

^{−}

^{1}···

*a*

^{−}

^{1}=

*a*

^{−}^{n}To prove Theorem 1 completely, we need to check the other cases, where each of the integers *m* and *n* is allowed to be zero or negative. This routine case-by-case verification is left to the student as __Exercise A__ at the end of this chapter.

In order to delve more deeply into the behavior of exponents we must use an elementary but very important property of the integers: From elementary arithmetic we know that we can divide any integer by any positive integer to get an integer quotient and an integer remainder. The remainder is nonnegative and less than the dividend. For example, 25 may be divided by 8, giving a quotient of 3, and leaving a remainder of 1:

Similarly, −25 may be divided by 8, with a quotient of −4 and a remainder of 7:

This important principle is called the division algorithm. In its precise form, it may be stated as follows:

**Theorem 2: Division algorithm** *if m and n are integers and n is positive, there exist unique integers q and r such that*

*m* = *nq* + *r* and 0 ≥ *r* < *n*

We call *q* the *quotient*, and *r* the *remainder*, in the division of *m* by *nx*.

At this stage we will take the division algorithm to be a *postulate* of the system of the integers. Later, in __Chapter 19__, we will turn things around, starting with a simpler premise and proving the division algorithm from it.

Let *G* be a group, and *a* an element of *G*. Let us observe that

*If there exists a nonzero integer m such that a ^{m}* =

*e, then there exists a positive integer n such that a*=

^{n}*e*.

Indeed, if *a ^{m}* =

*e*where

*m*is negative, then

*a*= (

^{−}^{m}*a*)

^{m}^{−}

^{1}=

*e*

^{−}

^{1}=

*e*. Thus,

*a*

^{−}

*=*

^{m}*e*where −

*m*is positive. This simple observation is crucial in our next definition. Let

*G*be an arbitrary group, and

*a*an element of

*G*:

**Definition** *If there exists a nonzero integer m such that a ^{m} – e, then the order of the element a is defined to be the least positive integer n such that a^{n}* =

*e*.

*If there does not exist any nonzero integer m such that a ^{m}* =

*e, we say that a has order*

**infinity.**Thus, in any group *G*, every element has an order which is either a positive integer or infinity. If the order of *a* is a positive integer, we say that *a* has *finite order;* otherwise, *a* has *infinite order*. For example, let us glance at S_{3}, whose table appears on page 72. (It is customary to use exponential notation for composition: for instance, *β ∘ β* = *β*^{2}, *β ∘ β ∘ β* = *β*^{3}, and so on.) The order of *α* is 2, because *α*^{2} = *ε* and 2 is the smallest positive integer which satisfies that equation. The order of *β* is 3, because *β*^{3} = ε, and 3 is the lowest positive power of *β* equal to *ε*. It is clear, similarly, that γ has order 2, δ has order 3, and *κ* has order 2. What is the order of *ε*?

It is important to note that one speaks of *powers* of *a* only when the group’s operation is called multiplication. When we use additive notation, we speak of *multiples* of *a* instead of powers of *a*. The *positive multiples* of *a* are *a*, *a* + *a*, *a* + *a* + *a*, and so on, while the *negative multiples* of *a* are −*a*,(−*a*) + (−*a*), (−*a*) + (−*a*) + (−*a*), and so on. In _{6}, the number 2 has order 3, because 2 + 2 + 2 = 0, and no smaller multiple of 2 is equal to 0. Similarly, the order of 3 is 2, the order of 4 is 3, and the order of 5 is 6.

In , the number 2 has infinite order, because no nonzero multiple of 2 is equal to 0. As a matter of fact, in , every nonzero number has infinite order.

The main fact about the order of elements is given in the next two theorems. In each of the following theorems, *G* is an arbitrary group and *a* is any element of *G*.

**Theorem 3: Powers of a, if a has finite order**

*if the order of a is n, there are exactly n different powers of a, namely*,

*a*^{0}, *a*, *a*^{2}, *a*^{3}, …, *a ^{n}*

^{−}

^{1}

What this theorem asserts is that every positive or negative power of *a* is equal to one of the above, and the above are all different from one another.

Before going on, remember that the order of *a* in *n*, hence

*a ^{n}* =

*e*

and *n* is the *smallest* positive integer which satisfies this equation.

PROOF OF THEOREM 3: Let us begin by proving that every power of *a* is equal to one of the powers *a*^{0},*a*^{1},*a*^{2}, …, *a ^{n}*

^{−}

^{1}. Let

*a*be any power of

^{m}*a*. Use the division algorithm to divide

*m*by

*n*:

*m* = *nq* + *r*0 ≤ *r* < *n*

Then *a ^{m}* =

*a*

^{nq}^{ +r}= (

*a*= (

^{nq}a^{r}*a*)

^{n}*=*

^{q}a^{r}*e*=

^{q}a^{r}*a*

^{r}Thus, *a ^{m}* =

*a*, and

^{r}*r*is one of the integers 0,1, 2, …,

*n*− 1.

Next, we will prove that *a*^{0},*a*^{1},*a*^{2}, …, *a ^{n}*

^{−}

^{1}are all different, Suppose not; suppose

*a*=

^{r}*a*, where

^{s}*r*and

*s*are distinct integers between 0 and

*n*− 1. Either

*r*<

*s*or

*s*<

*r*, say

*s*<

*r*. Thus 0 ≤

*s*<

*r*<

*n*, and consequently,

0 < *r* − *s* < *n* (1)

However, this is impossible, because by __Equation (1)__, *r* − *s* is a positive integer less than *n*, whereas *n* (the order of *a*) is the *smallest* positive integer such that *a ^{n}* =

*e*.

This proves that we cannot have *a ^{r}* =

*a*where

^{s}*r*≠

*s*. Thus,

*a*

^{0},

*a*

^{1},

*a*

^{2}, …,

*a*

^{n}^{ }

^{−}

^{ 1}are all different! ■

**Theorem 4: Powers of a, if a has infinite order**

*If a has order infinity, then all the powers of a are different. That is, if r and s are distinct integers, then*

*a*≠

^{r}*a*.

^{s}PROOF: Let *r* and *s* be integers, and suppose *a ^{r}* =

*a*.

^{s}But *a* has order infinity, and this means that *a ^{m}* is not equal to

*e*for any integer

*m*expect 0. Thus,

*r*−

*s*= 0, so

*r*=

*s*. ■

This chapter concludes with a technical property of exponents, which is of tremendous importance in applications.

If *a* is an element of a group, the order of *a* is the *least positive integer n* such that *a ^{n}* =

*e*. But there are

*other*integers, say

*t*, such that

*a*=

*e*. How are they related to

*n*? The answer is very simple:

**Theorem 5** *Suppose an element a in a group has order n*. Then *a ^{t}* =

*e*iff

*t is a multiple of n*(“

*t is a multiple of n” means that t*=

*nq for some integer q*).

PROOF: If *t* = *nq*, then *a ^{t}* =

*a*= (

^{nq}*a*)

^{n}*=*

^{q}*e*=

^{q}*e*. Conversely, suppose

*a*=

^{t}*e*. Divide

*t*by

*n*using the division algorithm:

*t* = *nq* + *r*0 ≤ *r* < *n*

Then

*e* = *a ^{t}* =

*a*

^{nq}^{ + r}= (

*a*)

^{n}*=*

^{q}a^{r}*e*=

^{q}a^{r}*a*

^{r}Thus, *a ^{r}* =

*e*, where 0 ≤

*r*<

*n*. If

*r*≠0, then

*r*is a positive integer less than

*n*, whereas

*n*is the

*smallest*positive integer such that

*a*=

^{n}*e*. Thus

*r*= 0, and therefore

*t*=

*nq*. ■

If *a* is any element of a group, we will denote the order of *a* by

ord(*a*)

**EXERCISES**

**A. Laws of Exponents**

Let *G* be a group and *a* ∈ *G*.

# ** 1** Prove that

*a*=

^{m}a^{n}*a*

^{m}^{ + n}in the following cases:

(*a*)*m* = *0*

(*b*)*m <* 0 and *n* > 0

(*c*)*m <* 0 and *n* < 0

**2**Prove that (*a ^{m}*)

*=*

^{n}*a*in the following cases:

^{mn}(*a*)*m* = 0

(*b*)*n* = 0

(*c*)*m* < 0 and *n* > 0

(*d*)*m* > 0 and *n* < 0

(*e*)*m* < 0 and *n* < 0

**3**Prove that (*a ^{n}*)

^{−}

^{ 1}=

*a*

^{−}

^{ n}in the following cases:

(*a*)*n* = 0

(*b*)*n* < 0

**B. Examples of Orders of Elements**

**1**What is the order of 10 in _{25}?

**2**What is the order of 6 in _{16}?

# ** 3**What is the order of

in *S*_{6}

**4**What is the order of 1 in *? What is the order of 1 in ?

**5**If *A* is the set of all the real numbers *x* ≠ 0,1,2, what is the order of

in *S _{A}*?

**6**Can an element of an *infinite* group have *finite* order? Explain.

**7**In _{24}, list all the elements (*a*) of order 2; (*b*) of order 3; (*c*) of order 4; (*d*) of order 6.

**C. Elementary Properties of Order**

Let *a*, *b*, and *c* be elements of a group *G*. Prove the following:

**1**Ord(*a*) = 1iff*a* = *e*.

**2**If ord(*a*) = *n*, then *a ^{n}*

^{ }

^{−}

^{ r}= (

*a*)

^{r}^{−}

^{1}.

**3**If *a ^{k}* =

*e*where

*k*is odd, then the order of

*a*is odd.

# ** 4** Ord(

*a*) = ord(

*bab*

^{ }

^{−}

^{1}).

**5**The order of *a*^{ }^{−}^{l} is the same as the order of *a*.

**6**The order of *ab* is the same as the order of *ba*. [HINT: if

then *a* is the inverse of *x*. Thus, *ax* = *e*.]

**7**Ord(*abc*) = ord(*cab*) = ord(*bca*).

**8**Let *x* = *a*^{1}*a*^{2} ··· *a ^{n}*, and let

*y*be a product of the same factors, permuted cyclically. (That is,

*y*=

*a*

_{k}a_{k}_{ + 1}···

*a*

_{n}a_{1}···

*a*

_{k}_{ }

_{−}

_{1}.) Then ord(

*x*) = ord(

*y*).

**D. Further Properties of Order**

Let *a* be any element of finite order of a group *G*. Prove the following:

**1**If *a ^{p}* =

*e*where

*p*is a prime number, then

*a*has order

*p*. (

*a*≠

*e*.)

# ** 2**The order of

*a*is a divisor (factor) of the order of

^{k}*a*.

**3**If ord(*a*) = *km*, then ord(*a ^{k}*) =

*m*.

**4**If ord(*a*) = *n* where *n* is odd, then ord(*a*^{2}) = *n*.

**5** If *a* has order *n*, and *a ^{r}* =

*a*, then

^{s}*n*is a factor of

*r*−

*s*.

**6**If *a* is the *only* element of order *k* in *G*, then *a* is in the center of *G*. (HINT: Use __Exercise C4__. Also, see __Chapter 4__, __Exercise C6__.)

**7**If the order of *a* is not a multiple of ra, then the order of *a ^{k}* is not a multiple of

*m*. (HINT: Use part 2.)

**8**If ord(*a*) = *mk* and *a ^{rk}* =

*e*, then

*r*is a multiple of

*m*.

**† E. Relationship between ord ( ab), ord (a), and ord (b)**

Let *a* and *b* be elements of a group *G*. Let ord(*a*) = *m* and ord(*b*) = *n*; let lcm(*m*, *n*) denote the least common multiple of *m* and *n*. Prove parts 1–5:

**1** If *a* and *b* commute, then *ord*(*ab*) is a divisor of lcm(*m*, *n*).

**2** If *m* and *n* are relatively prime, then no power of *a* can be equal to any power of *b* (except for *e*). (REMARK: Two integers are said to be relatively prime if they have no common factors except ±1.) (HINT: Use __Exercise D2__.)

**3** If *m* and *n* are relatively prime, then the products *a ^{i}b^{j}* (0 ≤

*i*<

*m*, 0 ≤

*j*<

*n*) are all distinct.

**4** Let *a* and *b* commute. If *m* and *n* are relatively prime, then *ord*(*ab*) = *mn*. (HINT: Use part 2.)

**5** Let *a* and *b* commute. There is an element *c* in *G* whose order is lcm(*m*, *n*). (HINT: Use part 4, above, together with __Exercise D3__. Let *c* = *a ^{i}b* where

*a*is a certain power of

^{i}*a*.)

**6** Give an example to show that part 1 is not true if *a* and *b* do not commute.

Thus, there is no simple relationship between ord(*ab*), ord(*a*), and ord(*b*) if *a* and *b* fail to commute.

**† F. Orders of Powers of Elements**

Let *a* be an element of order 12 in a group *G*.

**1** What is the smallest positive integer *k* such that *a*^{8k} = *e*? (HINT: Use Theorem 5 and explain carefully!)

# ** 2** What is the order of

*a*

^{8}?

**3** What are the orders of *a*^{9}, *a*^{10}, *a*^{5}?

**4** Which powers of *a* have the same order as *a*? [That is, for what values of *k* is ord(*a ^{k}*) = 12?]

**5** Let *a* be an element of order *m* in any group *G*. What is the order of *a ^{k}*? (Look at the preceding examples, and generalize. Do not prove.)

**6** Let *a* be an element of order *m* in any group *G*. For what values of *k* is ord(*a ^{k}*) =

*m*? (Look at the preceding examples. Do not prove.)

**† G. Relationship between ord( a) and ord(a^{k})**

From elementary arithmetic we know that every integer may be written uniquely as a product of prime numbers. Two integers *m* and *n* are said to be *relatively prime* if they have no prime factors in common. (For example, 15 and 8 are relatively prime.) Here is a useful fact: If *m* and *n* are relatively prime, and m is a factor of *nk*, then *m* is a factor of *k*. (Indeed, all the prime factors of *m* are factors of *nk* but not of *n*, hence are factors of *k*.)

Let *a* be an element of order *n* in a group *G*.

**1** Prove that if *m* and *n* are relatively prime, then *a ^{m}* has order

*n*. (HINT: If

*a*=

^{mk}*e*, use Theorem 5 and explain why

*n*must be a factor of

*k*.)

**2** Prove that if *a ^{m}* has order

*n*, then

*m*and

*n*are relatively prime. [HINT: Assume

*m*and

*n*have a common factor

*q*>1, hence we can write

*m*=

*m′ q*and

*n*=

*n′q*. Explain why (

*a*)

^{m}*=*

^{n}*e*, and proceed from there.]

**3** Let *l* be the least common multiple of *m* and *n*. Let *l*/*m* = *k*. Explain why (*a ^{m}*)

*=*

^{k}*e*.

**4** Prove: If (*a ^{m}*)

*=*

^{t}*e*, then

*n*is a factor of

*mt*. (Thus,

*mt*is a common multiple of

*m*and

*n*.) Conclude that

**5** Use parts 3 and 4 to prove that the order of *a ^{m}* is [lcm(

*m*,

*n*)]/

*m*.

**† H. Relationship between the Order of a and the Order of any kth Root of a**

Let *a* denote an element of a group *G*.

**1** Let *a* have order 12. Prove that if *a* has a cube root, say *a* = *b*^{3} for some *b* ∈ *G*, then *b* > has order 36. {HINT: Show that *b*^{36} = *e*; then show that for each factor *k* of 36, *b ^{k}* =

*e*is impossible. [Example: If

*b*

^{12}=

*e*, then

*b*

^{12}= (

*b*

^{3})

^{4}=

*a*

^{4}=

*e*.] Derive your conclusion from these facts.}

# ** 2** Let

*a*have order 6. If

*a*has a fourth root in

*G*, say

*a*=

*b*

^{4}, what is the order of

*b*?

**3** Let *a* have order 10. If *a* has a sixth root in *G*, say *a* = *b*^{6}, what is the order of *b*?

**4** Let *a* have order *n*, and suppose *a* has a sixth root in *G*, say *a* = *b ^{k}*. Explain why the order of

*b*is a factor of

*nk*.

Let

**5** Prove that *n* and *l* are relatively prime. [HINT: Suppose *n* and *l* have a common factor *q* > 1. Then *n* = *qn′* and *l* = *ql′*, so

Thus *b ^{n,k}* =

*e*. (Why?) Draw your conclusion from these facts.]

Thus, if *a* has order *n* and *a* has a *k*th root *b*, then *b* has order *nk*/*l*, where *n* and *l* are relatively prime.

**6** Let *a* have order *n*. Let *k* be an integer such that every prime factor of *k* is a factor of *n*. Prove: If *a* has a *k*th root *b*, then ord(*b*) = *nk*.