﻿ ﻿THE FUNDAMENTAL HOMOMORPHISM THEOREM - A Book of Abstract Algebra

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

### Chapter 16. THE FUNDAMENTAL HOMOMORPHISM THEOREM

Let G be any group. In Chapter 15 we saw that every quotient group of G is a homomorphic image of G. Now we will see that, conversely, every homomorphic image of G is a quotient group of G. More exactly, every homomorphic image of G is isomorphic to a quotient group of G.

It will follow that, for any groups G and H, H is a homomorphic image of G iff H is (or is isomorphic to) a quotient group of G. Therefore, the notions of homomorphic image and of quotient group are interchangeable.

The thread of our reasoning begins with a simple theorem.

Theorem 1 Let f: GH be a homomorphism with kernel K. Then

f(a) = f(b)iffKa = Kb

(In other words, any two elements a and b in G have the same image under f iff they are in the same coset of K.)

Indeed, What does this theorem really tell us? It says that if f is a homomorphism from G to H with kernel K, then all the elements in any fixed coset of K have the same image, and, conversely, elements which have the same image are in the same coset of K. It is therefore clear already that there is a one-to-one correspondence matching cosets of K with elements in H. It remains only to show that this correspondence is an isomorphism. But first, how exactly does this correspondence match up specific cosets of K with specific elements of H? Clearly, for each x, the coset Kx is matched with the element f(x). Once this is understood, the next theorem is easy.

Theorem 2 Let f: GH be a homomorphism of G onto H. If K is the kernel of f, then

HG/K

PROOF: To show that G/K is isomorphic to H, we must look for an isomorphism from G/K to H. We have just seen that there is a function from G/K to H which matches each coset Kx with the element f(x); call this function ϕ. Thus, ϕ is defined by the identity

ϕ(Kx) = f(x)

This definition does not make it obvious that ϕ(Kx) is uniquely defined. (If it is not, then we cannot properly call ϕ a function.) We must make sure that if Ka is the same coset as Kb, then ϕ(Ka) is the same as ϕ(Kb): that is,

if Ka = Kb then f(a) = f(b)

As a matter of fact, this is true by Theorem 1.

Now, let us show that ϕ is an isomorphism:

ϕ is injective: If ϕ(Ka) = ϕ(Kb then f(a) = f(b); so by Theorem 1, Ka = Kb.

ϕ is surjective, because every element of H is of the form f(x) = ϕ(Kx). Finally, ϕ(Ka · Kb) = ϕ(Kab) = f(ab) = f(a)f(b) = ϕ(Ka)ϕ(Kb).

Thus, ϕ is an isomorphism from G/K onto H. ■

Theorem 2 is often called the fundamental homomorphism theorem. It asserts that every homomorphic image of G is isomorphic to a quotient group of G. Which specific quotient group of G? Well, if f is a homomorphism from G onto H, then H is isomorphic to the quotient group of G by the kernel of f

The fact that f is a homomorphism from G onto H may be symbolized by writing Furthermore, the fact that K is the kernel of this homomorphism may be indicated by writing Thus, in capsule form, the fundamental homomorphism theorem says that Let us see a few examples:

We saw in the opening paragraph of Chapter 14 that is a homomorphism from 6 onto 3. Visibly, the kernel of f is {0, 3}, which is the subgroup of 6 generated by ⟨3⟩, that is, the subgroup (3). This situation may be symbolized by writing We conclude by Theorem 2 that 3 6/⟨3⟩

For another kind of example, let G and H be any groups and consider their direct product G × H.Remember that G × H consists of all the ordered pairs (x, y) as x ranges over G and y ranges over H. You multiply ordered pairs by multiplying corresponding components; that is, the operation on G × H is given by

(a, b) · (c, d) = (ac, bd)

Now, let f be the function from G × H onto H given by

f(x, y) = y

It is easy to check that f is a homomorphism. Furthermore, (x, y) is in the kernel of f iff f (x, y) = y = e. This means that the kernel of f consists of all the ordered pairs whose second component is e. Call this kernel G*; then

G* = {(x, e): xG}

We symbolize all this by writing By the fundamental homomorphism theorem, we deduce that H ≅ (G × H)/G*.[It is easy to see that G* is an isomorphic copy of G; thus, identifying G* with G, we have shown that, roughly speaking, (G × H)/GH.]

Other uses of the fundamental homomorphism theorem are given in the exercises.

EXERCISES

In the exercises which follow, FHT will be used as an abbreviation for fundamental homomorphism theorem.

A. Examples of the FHT Applied to Finite Groups

In each of the following, use the fundamental homomorphism theorem to prove that the two given groups are isomorphic. Then display their tables.

Example 2 and 6/⟨2⟩. is a homomorphism from 6 onto 2.(Do not prove that f is a homomorphism.) The kernel of f is {0, 2, 4} = ⟨2⟩. Thus, It follows by the FHT that 2 6/⟨2⟩.

1 5 and 20/⟨5⟩.

2 3 and 6/⟨3⟩.

3 2 and S3/{ε, β, δ}.

4 P2 and P3/K, where K = {0, {c}}. [HINT: Consider the function f(C) = C ∩ {a, b}. P3 is the group of subsets of {a, b, c}, and P2 of {a, b}.]

5 3 and ( 3 × 3)/K, where K = {(0, 0), (1, 1), (2, 2)}.[HINT: Consider the function f(a, b) = ab from 3 × 3 to 3.]

B. Example of the FHT Applied to ( )

Let α: ( ) → be defined by α(f) = f(l) and let β: ( ) → be defined by β(f) = f(2).

1 Prove that α and β are homomorphisms from ( ) onto .

2 Let J be the set of all the functions from to whose graph passes through the point (1, 0) and let K be the set of all the functions whose graph passes through (2, 0).Use the FHT to prove that  ( )/J and  ( )/K

3 Conclude that ( )/J ( )/K

C. Example of the FHT Applied to Abelian Groups

Let G be an abelian group. Let H = {x2: xG} and K = {x ∈ G: x2 = e}.

1 Prove that f(x) = x2 is a homomorphism of G onto H.

2 Find the kernel of f.

3 Use the FHT to conclude that HG/K

† D. Group of Inner Automorphisms of a Group G

Let G be a group. By an automorphism of G we mean an isomorphism f: GG.

# 1 The symbol Aut(G) is used to designate the set of all the automorphisms of G.Prove that the set Aut (G), with the operation ∘ of composition, is a group by proving that Aut(G) is a subgroup of SG.

2 By an inner automorphism of G we mean any function ϕa of the following form:

for every xa(x) = axa1

Prove that every inner automorphism of G is an automorphism of G.

3 Prove that, for arbitrary a, bG.

ϕaϕb = ϕaband(ϕa)1 = ϕa 1

4 Let I(G) designate the set of all the inner automorphisms of G. That is, I(G) = {ϕa: aG}.Use part 3 to prove that I(G) is a subgroup of Aut(G).Explain why I(G) is a group.

5 By the center of G we mean the set of all those elements of G which commute with every element of G, that is, the set C defined by

C = {aG: ax = xa for every xG}

Prove that aC if and only if axa1 = x for every xG.

6 Let h: GI(G) be the function defined by h(a) = ϕa. Prove that h is a homomorphism from G onto I(G) and that C is its kernel.

7 Use the FHT to conclude that I(G) is isomorphic with G/C.

† E.The FHT Applied to Direct Products of Groups

Let G and H be groups. Suppose J is a normal subgroup of G and K is a normal subgroup of H.

1 Show that the function f(x, y) = (Jx, Ky) is a homomorphism from G × H onto (G/J) × (H/K).

2 Find the kernel of f.

3 Use the FHT to conclude that (G × H)/(J ×K) ≅ (G/J) × (H/K).

† F. First Isomorphism Theorem

Let G be a group; let H and K be subgroups of G, with H a normal subgroup of G. Prove the following:

1 H K is a normal subgroup of K

# 2 If HK = {xy: xH and yK}, then HK is a subgroup of G.

3 H is a normal subgroup of HK.

4 Every member of the quotient group HK/H may be written in the form Hk for some kK.

5 The function f(k) = Hk is a homomorphism from K onto HK/H, and its kernel is H K

6 By the FHT, K/(H K) ≅ HK/H. (This is referred to as the first isomorphism theorem.)

† G. A Sharper Cayley Theorem

If H is a subgroup of a group G, let X designate the set of all the left cosets of H in G. For each element aG, define pa: XX as follows:

pa(xH) = (ax)H

1 Prove that each pa is a permutation of X.

2 Prove that h: GSX defined by h(a) = pa is a homomorphism.

# 3 Prove that the set {aH: xax1H for every xG}, that is, the set of all the elements of H whose conjugates are all in H, is the kernel of h.

4 Prove that if H contains no normal subgroup of G except {e}, then G is isomorphic to a subgroup of SX.

† H. Quotient Groups Isomorphic to the Circle Group

Every complex number a + bi may be represented as a point in the complex plane. The unit circle in the complex plane consists of all the complex numbers whose distance from the origin is 1; thus, clearly, the unit circle consists of all the complex numbers which can be written in the form

cos x + i sin x

for some real number x.

# 1 For each x , it is conventional to write cis x = cos x + i sin x. Prove that eis (x + y) = (cis x)(cis y).

2 Let T designate the set {cis x: x }, that is, the set of all the complex numbers lying on the unit circle, with the operation of multiplication. Use part 1 to prove that T is a group. (T is called the circle group.)

3 Prove that f(x) = cis x is a homomorphism from onto T.

4 Prove that ker f = {2: n } = ⟨2π⟩.

5 Use the FHT to conclude that T /⟨2⟩.

6 Prove that g(x) = cis 2πx is a homomorphism from onto T, with kernel .

7 Conclude that T / .

† I. The Second Isomorphism Theorem

Let H and K be normal subgroups of a group G, with H k Define ϕ: G/HG/K by ϕ(Ha) = Ka.

Prove parts 1–4:

1 ϕ is a well-defined function. [That is, if Ha = Hb, then ϕ(Ha) = ϕ(Hb).]

2 ϕ is a homomorphism.

3 ϕ is surjective.

4 ker ϕ K/H

5 Conclude (using the FHT) that (G/H)K/H) ≅ G /K.

† J. The Correspondence Theorem

Let f be a homomorphism from G onto H with kernel K: If S is any subgroup of H, let S* = {x ∈G: f(x)∈ S}. Prove:

1 S* is a subgroup of G.

2 K S*.

3 Let g be the restriction of f to S.*[That is, g(x) = f(x) for every x ∈ S*, and S* is the domain of g.] Then g is a homomorphism from S* onto S, and K = ker g.

4 SS*/K.

† K. Cauchy’s Theorem

Prerequisites: Chapter 13, Exercise I, and Chapter 15, Exercises G and H.

If G is a group and p is any prime divisor of |G|, it will be shown here that G has at least one element of order p. This has already been shown for abelian groups in Chapter 15, Exercise H4. Thus, assume here that G is not abelian. The argument will proceed by induction; thus, let |G| = k, and assume our claim is true for any group of order less than k. Let C be the center of G, let Ca be the centralizer of a for each aG, and let k = c + ks + ⋯ + kt be the class equation of G, as in Chapter 15, Exercise G2.

1 Prove: If p is a factor of |Ca| for any aG, where aC, we are done. (Explain why.)

2 Prove that for any aC in G, if p is not a factor of |Ca|, then p is a factor of (G: Ca).

3 Solving the equation k = c + ks + ⋯ + kt for c, explain why p is a factor of c. We are now done. (Explain why.)

† L. Subgroups of p-Groups (Prelude to Sylow)

Prerequisites: Exercise J; Chapter 15, Exercises G and H.

Let pbea prime number. A p-group is any group whose order is a power of p. It will be shown here that if |G| = pk then G has a normal subgroup of order pm for every m between 1 and k. The proof is by induction on |G|; we therefore assume our result is true for all /^-groups smaller than G. Prove parts 1 and 2:

1 There is an element a in the center of G such that ord (a) = p. (See Chapter 15, Exercises G and H.)

2a⟩ is a normal subgroup of G.

3 Explain why it may be assumed that G/⟨a⟩ has a normal subgroup of order pm 1.

# 4 Use Exercise J4 to prove that G has a normal subgroup of order pm.

SUPPLEMENTARY EXERCISES

Exercise sets M through Q are included as a challenge for the ambitious reader. Two important results of group theory are proved in these exercises: one is called Sylow’s theorem, the other is called the basis theorem of finite abelian groups.

† M. p-Sylow Subgroups

Prerequisites: Exercises J and K of this Chapter, Exercise I1 of Chapter 14, and Exercise D3 of Chapter 15.

Let p be a prime number. A finite group G is called a p-group if the order of every element x in G is a power p. (The orders of different elements may be different powers of p.) If H is a subgroup of any finite group G, and H is a p-group, we call H a p-subgroip of G. Finally, if K is a p-subgroup of G, and K is maximal (in the sense that K is not contained in any larger p-subgroup of G), then K is called a p-Sylow subgroup of G.

1 Prove that the order of any p-group is a power of p.(HINT:Use Exercise K.)

2 Prove that every conjugate of a p-Sylow subgroup of G is a p-Sylow subgroup of G.

Let K be a p-Sylow subgroup of G, and N = N(K) the normalizer of K.

3 Let aN, and suppose the order of Ka in N/K is a power of p. Let S = ⟨Ka⟩ be the cyclic subgroup of N/K generated by Ka. Prove that N has a subgroup S* such that S*/K is a p-group. (HINT: See Exercise J4.)

4 Prove that S* is a p-subgroup of G (use Exercise D3, Chapter 15). Then explain why S* = K, and why it follows that Ka = K.

5 Use parts 3 and 4 to prove: no element of N/K has order a power of p (except, trivially, the identity element).

6 If aN and the order of p is a power of p, then the order of Ka (in N/K) is also a power of p. (Why?) Thus, Ka = K.(Why?)

7 Use part 6 to prove: if aKal = K and the order of a is a power of p, then aK.

† N. Sylow’s Theorem

Prerequisites: Exercises K and M of this Chapter and Exercise I of Chapter 14.

Let G be a finite group, and K a p-Sylow subgroup of G.Let X be the set of all the conjugates of K. See Exercise M2. If C1, C2X, let C1C2 iff C1 = aC2al for some αK

1 Prove that ∼ is an equivalence relation on X.

Thus, ∼ partitions X into equivalence classes. If C ∈, X let the equivalence class of C be denoted by [C]. 2 For each CX, prove that the number of elements in [C] is a divisor of |K|. (HINT: Use Exercise I10 of Chapter 14.) Conclude that for each CX, the number of elements in [C] is either 1 or a power of p.

3 Use Exercise M7 to prove that the only class with a single element is [K],

4 Use parts 2 and 3 to prove that the number of elements in X is kp + 1, for some integer k.

5 Use part 4 to prove that (G:N) is not a multiple of p.

6 Prove that (N: K) is not a multiple of p. (Use Exercises K and M5.)

7 Use parts 5 and 6 to prove that (G: K) is not a multiple of p.

8 Conclude: Let G be a finite group of order pkm, where p is not a factor of m. Every p-Sylow subgroup K of G has order pk.

Combining part 8 with Exercise L gives

Let G be a finite group and let p be a prime number. For each n such that pn divides |G|, G has a subgroup of order pn.

This is known as Sylow’s theorem.

† O. Lifting Elements from Cosets

The purpose of this exercise is to prove a property of cosets which is needed in Exercise Q. Let G be a finite abelian group, and let a be an element of G such that ord(a) is a multiple of ord(x) for every xG. Let H = ⟨a⟩. We will prove:

For every xG, there is some yG such that Hx = Hy and ord(y) = ord(Hy).

This means that every coset of H contains an element y whose order is the same as the coset’s order.

Let x be any element in G, and let ord (a) = t, ord(x) = s, and ord (Hx) = r.

1 Explain why r is the least positive integer such that xr equals some power of a, say xr = am.

2 Deduce from our hypotheses that r divides s, and s divides t.

Thus, we may write s = ru and t = , so in particular, t = ruυ.

3 Explain why amu = e, and why it follows that mu = tz for some integer z. Then explain why m = rυz.

4 Setting y = xaυz, prove that Hx = Hy and ord(y) = r, as required.

† P. Decomposition of a Finite Abelian Group into p-Groups

Let G be an abelian group of order pkm, where pk and m are relatively prime (that is, pk and m have no common factors except ±1). (REMARK: If two integers j and k are relatively prime, then there are integers s and t such that sj + tk= 1. This is proved on page 220.)

Let Gpk be the subgroup of G consisting of all elements whose order divides pk. Let Gm be the subgroup of G consisting of all elements whose order divides ra. Prove:

1 For any xG and integers s and t, xspkGm and xtmGpk.

2 For every xG, there are yGpk and zGm such that x = yz.

3 Gpk Gm = {e}.

4 GGpk × Gm. (See Exercise H, Chapter 14.)

5 Suppose |G| has the following factorization into primes: . Then GG1 × G2 × ⋯ × Gn where for each i = 1, …, n, Gi is a pi-group.

Q. Basis Theorem for Finite Abelian Groups

Prerequisite: Exercise P.

As a provisional definition, let us call a finite abelian group “decomposable” if there are elements al, …, anG such that:

(Dl) For every xG, there are integers k1, …, kn such that . (D2) If there are integers l1, …, ln such that then  .

If (Dl) and (D2) hold, we will write G = [a1, a2, …, an]. Assume this in parts 1 and 2.

1 Let G′ be the set of all products range over . Prove that G′ is a subgroup of G, and G′ = [a2, …, an].

2 Prove: G ≅ ⟨a1⟩ × G′. Conclude that G ≅⟨a1⟩ × ⟨a2⟩ × ⋯ × ⟨an

In the remaining exercises of this set, let p be a prime number, and assume G is a finite abelian group such that the order of every element in G is some power of p. Let aG be an element whose order is the highest possible in G. We will argue by induction to prove that G is “decomposable.” Let H = ⟨a⟩.

3 Explain why we may assume that G/H = [Hb1, …, Hbn] for some bl, …, bnG.

By Exercise O, we may assume that for each i = 1, …, n, ord (bi) = (Hbi). We will show that G = [a, b1, …, bn].

4 Prove that for every xG, there are integers k0, k1, …, kn such that 5 Prove that if , then . Conclude that G = [a, b,1, …, bn].

6 Use Exercise P5, together with parts 2 and 5 above, to prove: Every finite abelian group G is a direct product of cyclic groups of prime power order. (This is called the basis theorem of finite abelian groups.)

It can be proved that the above decomposition of a finite abelian group into cyclic p-groups is unique, except for the order of the factors. We leave it to the ambitious reader to supply the proof of uniqueness.

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