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 = S₃ 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 S₃. The reason is that, although H is a subgroup of S₃, H is not a normal subgroup of S₃. 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 = h₁c for some h₁ ∈ H. If Hb = Hd, then b ∈ Hd; hence b = h₂d from some h₂ ∈ H. Thus,

ab = h₁ch₂d = h₁(ch₂)d

But ch₂ ∈ cH = Hc (the last equality is true by Theorem 1). Thus, ch₂ = h₃c for some h₃ ∈ H. Returning to ab,

ab = h₁(ch₂)d = h₁(h₃c)d = (h₁h₃)(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, _n is a homomorphic image of .

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^m has finite order, say 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⁻¹b⁻¹ where a and b are in G. The reason such a product is called a commutator is that

aba⁻¹b⁻¹ = e iff ab = ba

In other words, aba⁻¹b⁻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⁻¹y⁻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 ≅ Z₅.)

2 G = S₃, H = {ε, β, δ}.

3 G = D₄, H = {R₀, R₂}. (See page 73.)

4 G = D₄, H = {R₀, R₂, R₄, R₅}.

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

6 G = P₃, H = {ø, {1}}. (P₃ 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 = 2x}

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 x² ∈ 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 x² ∈ 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 xⁿ ∈ 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 xⁿ ∈ H.

# 4 Every element of G/H has a square root iff for every x ∈ G, there is some y ∈ G such that xy² ∈ 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 xaⁿ ∈ 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_p is a subgroup of G. Prove that G/H_p has no elements whose order is a nonzero power of 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′aⁿ, and remember that 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⁻¹ax). 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 S₁, S₂,. . ., S_t be the distinct conjugacy classes of G, and let k₁, k₂,.. ., k_t be their sizes. Then |G| = k₁ + k₂ + … + k_t (This is called the 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_t, where k_s,. . ., k_t are the sizes of all the distinct conjugacy classes of elements x ∉ C.

3 For each i ∈ {s, s + 1,..., t}, k_i is equal to a power of p. (See Chapter 13, Exercise I6.)

4 Solving the equation |G| = c + k_s + · · · + k_t for 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| = p², G must be abelian. (Use the preceding Exercise F.)

# 6 Prove: If |G| = p², then either G ≅ _p2 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.)