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

### Chapter 15. QUOTIENT GROUPS

In __Chapter 14__ we learned to recognize when a group *H* is a homomorphic image of a group *G*. Now we will make a great leap forward by learning a method for actually *constructing all the homomorphic images of any group*. This is a remarkable procedure, of great importance in algebra. In many cases this construction will allow us to deliberately select *which* properties of a group *G* we wish to preserve in a homomorphic image, and which other properties we wish to discard.

The most important instrument to be used in this construction is the notion of a normal subgroup. Remember that a *normal* subgroup of *G* is any subgroup of *G* which is closed with respect to conjugates. We begin by giving an elementary property of normal subgroups.

**Theorem 1** *If H is a normal subgroup of G, then aH* = *Ha for every a ∈ G*.

(In other words, there is no distinction between left and right cosets for a normal subgroup.)

PROOF: Indeed, if *x* is any element of *aH*, then *x* = *ah* for some *h* ∈ *H*. But *H* is closed with respect to conjugates; hence *aha*^{−}^{1} ∈ *H*. Thus, *x* = *ah* = (*aha*^{−}^{1})*a* is an element of *Ha*. This shows that every element of *aH* is in *Ha*; analogously, every element of *Ha* is in *aH*. Thus, *aH* = *Ha*. ■

Let *G* be a group and let *H* be a subgroup of *G*. There is a way of combining cosets, called *coset multiplication*, which works as follows: *the coset of a, multiplied by the coset of b, is defined to be the coset of ab*. In symbols,

*Ha* · *Hb* = *H*(*ab*)

This definition is deceptively simple, for it conceals a fundamental difficulty. Indeed, it is not at all clear that the product of two cosets *Ha* and *Hb*, multiplied together in this fashion, is *uniquely defined*. Remember that *Ha* may be the same coset as *Hc* (this happens iff *c* is in *Ha*), and, similarly, *Hb* may be the same coset as *Hd*. Therefore, the product *Ha* · *Hb* is the same as the product *He* · *Hd*. Yet it may easily happen that *H*(*ab*) is *not* the same coset as *H*(*cd*). Graphically,

For example, if *G* = *S*_{3} and *H* = {*ε, α*}, then

and yet

*H*(*β ∘* *δ*) = *Hε* ≠ *Hβ* = *H*(*γ ∘* *κ*)

Thus, coset multiplication *does not work* as an operation on the cosets of *H* = {*ε, α*} in *S*_{3}. The reason is that, although *H* is a subgroup of *S*_{3}, *H* is *not a normal subgroup of S*_{3}. If *H* were a normal subgroup, coset multiplication would work. The next theorem states exactly that!

**Theorem 2** *Let H be a normal subgroup of G. If Ha* = *He and Hb* = *Hd, then H*(*ab*) = *H*(*cd*).

PROOF: If *Ha* = *Hc*, then *a* ∈ *Hc;* hence *a* = *h*_{1}*c* for some *h*_{1} ∈ *H*. If *Hb* = *Hd*, then *b* ∈ *Hd;* hence *b* = *h*_{2}*d* from some *h*_{2} ∈ *H*. Thus,

*ab* = *h*_{1}*ch*_{2}*d* = *h*_{1}(*ch*_{2})*d*

But *ch*_{2} ∈ *cH* = *Hc* (the last equality is true by __Theorem 1__). Thus, *ch*_{2} = *h*_{3}*c* for some *h*_{3} ∈ *H*. Returning to *ab*,

*ab* = *h*_{1}(*ch*_{2})*d* = *h*_{1}(*h*_{3}*c*)*d* = (*h*_{1}*h*_{3})(*cd*)

and this last element is clearly in *H*(*cd*).

We have shown that *ab* ∈ *H*(*cd*). Thus, by Property (1) in __Chapter 13__, *H*(*ab*) = *H*(*cd*). ■

We are now ready to proceed with the construction promised at the beginning of the chapter. Let *G* be a group and let *H* be a normal subgroup of *G*. Think of the set which consists of *all the cosets of H*. This set is conventionally denoted by the symbol *G*/*H*. Thus, if *Ha*, *Hb*, *He*,. . . are cosets of *H*, then

*G*/*H* = {*Ha*, *Hb*, *Hc*,. . .}

We have just seen that *coset multiplication* is a valid operation on this set. In fact,

**Theorem 3** *G/H with coset multiplication is a group*.

PROOF: Coset multiplication is associative, because

The identity element of *G*/*H* is *H* = *He*, for *Ha* · *He* = *Ha* and *He* · *Ha* = *Ha* for every coset *Ha*.

Finally, the inverse of any coset *Ha* is the coset *Ha*^{−}^{1}, because *Ha* · *Ha*^{−}^{1} = *Haa*^{−}^{1} = *He* and *Ha*^{−}^{1} · *Ha* = *Ha*^{−}^{1} *a* *He*.

The group *G*/*H* is called the *factor group*, or *quotient group* of *G* by *H*.

And now, the pièce de *résistance:*

**Theorem 4** *G/H is a homomorphic image of G*.

PROOF: The most obvious function from *G* to *G*/*H* is the function *f* which carries every element to its own coset, that is, the function given by

*f*(*x*) = *Hx*

This function is a homomorphism, because

*f*(*xy*) = *Hxy* = *Hx* · *Hy* = *f*(*x*)*f*(*y*)

*f* is called the *natural homomorphism* from *G* onto *G*/*H*. Since there is a homomorphism from *G* onto *G*/*H*, *G*/*H* is a homomorphic image of *G*. ■

Thus, when we construct quotient groups of *G*, we are, in fact, constructing homomorphic images of *G*. The quotient group construction is useful because it is a way of actually manufacturing homomorphic images of any group *G*. In fact, as we will soon see, it is a way of manufacturing *all* the homomorphic images of *G*.

Our first example is intended to clarify the details of quotient group construction. Let be the group of the integers, and let ⟨6⟩ be the cyclic subgroup of which consists of all the multiples of 6. Since is abelian, and every subgroup of an abelian group is normal, ⟨6⟩ is a normal subgroup of . Therefore, we may form the quotient group /⟨6⟩. The elements of this quotient group are all the cosets of the subgroup ⟨6⟩, namely:

These are *all* the different cosets of ⟨6⟩, for it is easy to see that ⟨6⟩ + 6 = ⟨6⟩ + 0, ⟨6⟩ + 7 = ⟨6⟩ + 1, ⟨6⟩ + 8 = ⟨6⟩ + 2, and so on.

Now, the operation on is denoted by +, and therefore we will call the operation on the cosets *coset addition* rather than coset multiplication. But nothing is changed except the name; for example, the coset ⟨6⟩ + 1 added to the coset ⟨6⟩ + 2 is the coset ⟨6⟩ + 3. The coset ⟨6⟩ + 3 added to the coset ⟨6⟩ + 4 is the coset ⟨6⟩ + 7, which is the same as ⟨6⟩ + 1. To simplify our notation, let us agree to write the cosets in the following shorter form:

Then /⟨6⟩ consists of the six elements and , and its operation is summarized in the following table:

The reader will perceive immediately the similarity between this group and _{6}. As a matter of fact, the quotient group construction of /⟨6⟩ is considered to be the rigorous way of constructing _{6}. So from now on, we will consider _{6} to be the same as /⟨6⟩; and, in general, we will consider * _{n}* to be the same as /⟨

*n*⟩. In particular, we can see that for any

*n*,

*is a homomorphic image of .*

_{n}Let us repeat: The motive for the quotient group construction is that it gives us a way of actually *producing* all the homomorphic images of any group *G*. However, what is even more fascinating about the quotient group construction is that, in practical instances, we can often choose *H* so as to “factor out” unwanted properties of *G*, and preserve in *G*/*H* only “desirable” traits. (By “desirable” we mean desirable within the context of some specific application or use.) Let us look at a few examples.

First, we will need two simple properties of cosets, which are given in the next theorem.

**Theorem 5** *Let G be a group and H a subgroup of G. Then*

(i)*Ha = Hb iff ab*^{−}^{1} ∈ *H and*

(ii)*Ha* = *H* *iff* *a* ∈ *H*

PROOF: If *Ha* = *Hb*, then *a* ∈ *Hb*, so *a* = *hb* for some *h* ∈ *H*. Thus,

*ab*^{−}^{1} = *h* ∈ *H*

If *ab*^{−}^{1} ∈ *H*, then *ab*^{−}^{1} = *h* for *h* ∈ *H*, and therefore *a* = *hb* ∈ *Hb*. It follows by Property (1) of __Chapter 13__ that *Ha* = *Hb*.

This proves (i). It follows that *Ha* = *He* iff *ae*^{−}^{1} = *a* ∈ *H*, which proves (ii). ■

For our first example, let *G* be an abelian group and let *H* consist of *all the elements of G which have finite order*. It is easy to show that *H* is a subgroup of *G*. (The details may be supplied by the reader.) Remember that in an abelian group every subgroup is normal; hence *H* is a normal subgroup of *G*, and therefore we may form the quotient group *G*/*H*. We will show next that in *G/H, no* element except the neutral element has finite order.

For suppose *G/H* has an element *Hx* of finite order. Since the neutral element of *G/H* is *H*, this means there is an integer *m* ≠ 0 such that (*Hx*)* ^{m}* =

*H*, that is,

*Hx*=

^{m}*H*. Therefore, by

__Theorem 5__(ii),

*x*∈

^{m}*H*, so

*x*has finite order, say

^{m}*t:*

(*x ^{m}*)

*=*

^{t}*x*=

^{mt}*e*

But then *x* has finite order, so *x* ∈ *H*. Thus, by __Theorem 5__(ii), *Hx* = *H*. This proves that in *G/H*, the only element *Hx* of finite order is the neutral element *H*.

Let us recapitulate: If *H* is the subgroup of *G* which consists of all the elements of *G* which have finite order, then in *G/H, no* element (except the neutral element) has finite order. Thus, in a sense, we have “*factored* *out”* *all the elements of finite order* (*they are all in H*) *and produced a* *quotient group GIH whose elements all have infinite order* (except for the neutral element, which necessarily has order 1).

Our next example may bring out this idea even more clearly. Let *G* be an arbitrary group; by a *commutator* of *G* we mean any element of the form *aba*^{−}^{1}*b*^{−}^{1} where *a* and *b* are in *G*. The reason such a product is called a commutator is that

*aba*^{−}^{1}*b*^{−}^{1} = *e* iff *ab* = *ba*

In other words, *aba*^{−}^{1}*b*^{−}^{1} reduces to the neutral element whenever *a* and *b* commute—and *only* in that case! Thus, in an abelian group all the commutators are equal to *e*. In a group which is not abelian, the number of distinct commutators may be regarded as a measure of the extent to which *G* departs from being commutative. (The fewer the commutators, the closer the group is to being an abelian group.)

We will see in a moment that if *H* is a subgroup of *G* which *contains all the commutators of G*, then *G*/*H* is abelian! What this means, in a fairly accurate sense, is that *when we factor out the commutators of G we get a quotient group which has no commutators* (except, trivially, the neutral element) *and which is therefore abelian*.

To say that *G/H* is abelian is to say that for any two elements *Hx* and *Hy* in *G/H, HxHy* = *HyHx;* that is, *Hxy* = *Hyx*. But by __Theorem 5__(ii),

*Hxy* = *Hyx* iff *xy*(*yx*)^{−}^{1} ∈ *H*

Now *xy*(*yx*)^{−}^{1} is the commutator *xyx*^{−}^{1}*y*^{−}^{1}; so if all commutators are in *H*, then *G/H* is abelian.

**EXERCISES**

**A. Examples of Finite Quotient Groups**

In each of the following, *G* is a group and *H* is a normal subgroup of *G*. List the elements of *G/H* and then write the table of *G/H*.

**Example** *G* = _{6} and *H* = {0, 3}

The elements of *G/H* are the three cosets *H* = *H* + 0 = {0, 3}, *H* + 1 = {1, 4}, and *H* + 2 = {2, 5}. (Note that *H* + 3 is the same as *H* + 0, *H* + 4 is the same as *H* + 1, and *H* + 5 is the same as *H* + 2.) The table of *G/H* is

**1** *G* = _{10}, *H* = {0,5}. (Explain why *G/H* ≅ Z_{5}.)

**2** *G* = *S*_{3}, *H* = {*ε*, *β*, *δ*}.

**3** *G* = *D*_{4}, *H* = {*R*_{0}, *R*_{2}}. (See page 73.)

**4** *G* = *D*_{4}, *H* = {*R*_{0}, *R*_{2}, *R*_{4}, *R*_{5}}.

**5** *G* = _{4} × _{2}, *H* = ⟨(0,1)⟩ = the subgroup of _{4} × _{2} generated by (0,1).

**6** *G* = *P*_{3}, *H* = {ø, {1}}. (*P*_{3} is the group of subsets of {1, 2, 3}.)

**B. Examples of Quotient Groups of × **

In each of the following, *H* is a subset of × .

(*a*) Prove that *H* is a normal subgroup of × . (Remember that every subgroup of an abelian group is normal.)

(*b*) In geometrical terms, describe the elements of the quotient group *G/H*.

(*c*) In geometrical terms or otherwise, describe the operation of *G/H*.

**1** *H* = {(*x*,0):*x* ∈ }

**2** *H* = {(*x*, *y*):*y* = −*x*}

**3** *H* = {(*x*, *y*):*y* = 2*x*}

**C. Relating Properties of H to Properties of G/H**

In parts 1-5 below, *G* is a group and if is a normal subgroup of *G*. Prove the following (__Theorem 5__ will play a crucial role):

**1** If *x*^{2} ∈ *H* for every *x* ∈ *G*, then every element of *G/H* is its own inverse. Conversely, if every element of *G/H* is its own inverse, then *x*^{2} ∈ *H* for all *x* ∈ *G*.

**2** Let *m* be a fixed integer. If *x ^{m}* ∈

*H*for every

*x*∈

*G*, then the order of every element in

*G/H*is a divisor of

*m*. Conversely, if the order of every element in

*G/H*is a divisor of

*m*, then

*x*∈

^{m}*H*for every

*x*∈

*G*.

**3** Suppose that for every *x* ∈ *G*, there is an integer *n* such that *x ^{n}* ∈

*H*; then every element of

*G/H*has finite order. Conversely, if every element of

*G/H*has finite order, then for every

*x*∈

*G*there is an integer

*n*such that

*x*∈

^{n}*H*.

**# 4** Every element of

*G/H*has a square root iff for every

*x*∈

*G*, there is some

*y*∈

*G*such that

*xy*

^{2}∈

*H*.

**5** *G/H* is cyclic iff there is an element *a* ∈ *G* with the following property: for every *x* ∈ *G*, there is some integer *n* such that *xa ^{n}* ∈

*H*.

**6** If *G* is an abelian group, let *H _{p}* be the set of all

*x*∈

*H*whose order is a power of

*p*. Prove that

*H*is a subgroup of

_{p}*G*. Prove that

*G/H*has no elements whose order is a nonzero power of

_{p}*p*

**7** (*a*) If *G/H* is abelian, prove that *H* contains all the commutators of *G*.

(*b*) Let *K* be a normal subgroup of *G*, and *H* a normal subgroup of *K*. If *G*/*H* is abelian, prove that *G/K* and *K/H* are both abelian.

**D. Properties of G Determined by Properties of G/H and H**

There are some group properties which, if they are true in *G/H* and in *H*, must be true in *G*. Here is a sampling. Let *G* be a group, and *H* a normal subgroup of *G*. Prove the following:

**1** If every element of *G/H* has finite order, and every element of *H* has finite order, then every element of *G* has finite order.

**2** If every element of *G/H* has a square root, and every element of *H* has a square root, then every element of G has a square root. (Assume G is abelian.)

**3** Let *p* be a prime number. If *G/H* and *H* are *p*-groups, then *G* is a *p*-group. A group *G* is called a *p-group* if the order every element *x* in *G* is a power of *p*.

**# 4** If

*G/H*and

*H*are finitely generated, then

*G*is finitely generated. (A group is said to be finitely generated if it is generated by a finite subset of its elements.)

**E. Order of Elements in Quotient Groups**

Let *G* be a group, and *H* a normal subgroup of *G*. Prove the following:

**1** For each element *a* ∈ *G*, the order of the element *Ha* in *G/H* is a divisor of the order of *a* in *G*. (HINT: Use __Chapter 14__, __Exercise F1__.)

**2** If (*G*: *H*) = *m*, the order of every element of *G/H* is a divisor of *m*.

**3** If (*G*: *H*) = *p*, where *p* is a prime, then the order of every element *a* ∉ *H* in *G* is a multiple of *p*. (Use part 1.)

**4** If *G* has a normal subgroup of index *p*, where *p* is a prime, then *G* has at least one element of order *p*.

**5** If (*G*: *H*) = *m*, then *a ^{m}* ∈

*G*for every

*a*∈

*G*.

**# 6** In /, every element has finite order.

**† F. Quotient of a Group by Its Center**

The *center* of a group *G* is the normal subgroup *C* of *G* consisting of all those elements of *G* which commute with every element of *G*. Suppose the quotient group *G/C* is a cyclic group; say it is generated by the element *Ca* of *G/C*. Prove parts 1-3:

**1** For every *x* ∈ *G*, there is some integer *m* such that *Cx* = *Ca ^{m}*.

**2** For every *x* ∈ *G*, there is some integer *m* such that *x* = *ca ^{m}*, where

*c*∈

*C*.

**3** For any two elements *x* and *y* in *G*, *xy* = *yx*. (HINT: Use part 2 to write *x* = *ca ^{m}*,

*y*=

*c′a*, and remember that

^{n}*c*,

*c′*∈

*C*.)

**4** Conclude that if *G/C* is cyclic, then *G* is abelian.

**† G. Using the Class Equation to Determine the Size of the Center**

{*Prerequisite: Chapter 13, Exercise I*.)

Let *G* be a finite group. Elements *a* and *b* in *G* are called *conjugates* of one another (in symbols, *a ~ b*) iff *a* = *xbx*^{−}^{1} for some *x* ∈ *G* (this is the same as *b* − *x*^{−}^{1}*ax*). The relation ~ is an equivalence relation in *G*; the equivalence class of any element *a* is called its *conjugacy class*. Hence *G* is partitioned into conjugacy classes (as shown in the diagram); the size of each conjugacy class divides the order of *G*. (For these facts, see __Chapter 13__, __Exercise I__.)

“Each element of the center *C* is alone in its conjugacy class.”

Let *S*_{1}, *S*_{2},. . ., *S _{t}* be the distinct conjugacy classes of

*G*, and let

*k*

_{1},

*k*

_{2},.. .,

*k*be their sizes. Then |

_{t}*G*| =

*k*

_{1}+

*k*

_{2}+ … +

*k*(This is called the

_{t}*class equation*of

*G*.)

Let *G* be a group whose order is a power of a prime *p*, say |*G*| = *p ^{k}*. Let

*C*denote the center of

*G*.

Prove parts 1-3:

**1** The conjugacy class of *a* contains *a* (and no other element) iff *a* ∈ *C*.

**2** Let *c* be the order of *C*. Then |*G*| = *c* + *k _{s}* +

*k*

_{s}_{ + 1}+ ··· +

*k*, where

_{t}*k*,. . .,

_{s}*k*are the sizes of all the distinct conjugacy classes of elements

_{t}*x*∉

*C*.

**3** For each *i* ∈ {*s*, *s* + 1,..., *t*}, *k _{i}* is equal to a power of

*p*. (See

__Chapter 13__,

__Exercise I__6.)

**4** Solving the equation |*G*| = *c* + *k _{s}* + · · · +

*k*for

_{t}*c*, explain why

*c*is a multiple of

*p*

We may conclude from part 4 that *C* must contain more than just the one element *e*; in fact, |*C*| is a multiple of *p*.

**5** Prove: If |*G*| = *p*^{2}, *G* must be abelian. (Use the preceding Exercise F.)

**# 6** Prove: If |

*G*| =

*p*

^{2}, then either

*G*≅

_{p}_{2}or

*G*≅ ∈

*× ∈*

_{p}*.*

_{p}**† H. Induction on | G|: An Example**

Many theorems of mathematics are of the form “*P*(*n*) is true for every positive integer *n*.” [Here, *P*(*n*) is used as a symbol to denote some statement involving *n*.] Such theorems can be proved by induction as follows:

(*a*) Show that *P*(*n*) is true for *n* = 1.

(*b*) For any fixed positive integer *k*, show that, if *P*(*n*) is true for every *n* < *k*, then *P*(*n*) must also be true for *n* = *k*.

If we can show (*a*) and (*b*), we may safely conclude that *P*(*n*) is true for all positive integers *n*.

Some theorems of algebra can be proved by induction on the order *n* of a group. Here is a classical example: Let *G* be a finite abelian group. We will show that G must contain at least one element of order *p*, for every prime factor *p* of |*G*|. If |*G*| = 1, this is true by default, since no prime *p* can be a factor of 1. Next, let |*G*| = *k*, and suppose our claim is true for every abelian group whose order is less than *k*. Let *p* be a prime factor of *k*.

Take any element *a* ≠ *e* in *G*. If ord(*a*) = *p* or a multiple of *p*, we are done!

**1** If ord(*a*) = *tp* (for some positive integer *t*), what element of *G* has order *p*?

**2** Suppose ord(*a*) *is not equal to a multiple of p*. Then *G*/⟨*a*⟩ is a group having fewer than *k* elements. (Explain why.) The order of *G*/⟨*a*⟩ is a multiple of *p*. (Explain why.)

**3** Why must *G*/(*a*) have an element of order *p*?

**4** Conclude that *G* has an element of order *p*. (HINT: Use Exercise El.)