Case 3. Suppose that N contains the pair of disjoint transpositions (ab) Ž (cd). Then, if e is the element of {1, 2, 3, 4, 5} not appearing in these transpositions, we have

(abe)−1 Ž (ab) Ž (cd) Ž (abe) = (aeb) Ž (ab) Ž (cd) Ž (abe) = (ae) Ž (cd) N.

Also, (ab) Ž (cd) Ž (ae) Ž (cd) = (aeb) N , and again we are back to case 1.

We have shown that any normal subgroup of A5 containing more than one element must be A5 itself.

A group without any proper normal subgroups is called a simple group. The term simple must be understood in the technical sense that it cannot be broken down, because it cannot have any nontrivial quotient groups. This is analogous to a prime number, which has no nontrivial quotients. Apart from the cyclic groups of prime order, which have no proper subgroups of any kind, simple groups are comparatively rare.

The group A5 is of great interest to mathematicians because it is used in Galois theory to show that there is an equation of the fifth degree that cannot be solved by any algebraic formula.

It can be shown that every alternating group An, n 5, is simple. The cyclic groups Cp , p a prime, are another infinite series of simple groups (the abelian ones), and other series have been known for decades. But it was not until 1981 that the finite simple groups were completely classified. This was the culmination of more than 30 years of effort by hundreds of mathematicians, yielding thousands of pages of published work, and was one of the great achievements of twentieth-century mathematics. One spectacular landmark came in 1963, when J. G. Thompson and W. Feit verified a long-standing conjecture of W. Burnside (1852–1927) that every finite, non-abelian, simple group has even order (the proof is more than 250 pages long!). The main difficulty in the classification was the existence of sporadic finite simple groups, not belonging to any of the known families. The largest of these, known as the monster, has order approximately 2 × 1053. The complete classification encompasses several infinite families and exactly 26 sporadic groups.

MORPHISM THEOREM

The morphism theorem is a basic result of group theory that describes the relationship between morphisms, normal subgroups, and quotient groups. There is an analogous result for most algebraic systems, including rings and vector spaces.

If f : G → H is a group morphism, the kernel of f , denoted by Kerf , is defined to be the set of elements of G that are mapped by f to the identity of H . That is, Kerf = {g G|f (g) = eH }.

Proposition 4.23. Let f : G → H be a group morphism. Then:

(i)Kerf is a normal subgroup of G.

(ii)f is injective if and only if Kerf = {eG}.

Proof. (i) We first show that Kerf is a subgroup of G. Let a, b Kerf so

Proposition 4.24. For any group morphism f : G → H , the image of f , Imf = {f (g)|g G}, is a subgroup of H (although not necessarily normal).

Theorem 4.25. Morphism Theorem for Groups. Let K be the kernel of the group morphism f : G → H . Then G/K is isomorphic to the image of f , and the isomorphism ψ: G/K → Imf is defined by ψ(Kg) = f (g).

This result is also known as the first isomorphism theorem; the second and third isomorphism theorems are given in Exercises 4.43 and 4.44.

Proof. The function ψ is defined on a coset by using one particular element in the coset, so we have to check that ψ is well defined; that is, it does not matter which element we use. If Kg = Kg , then g ≡ g mod K so g g−1 = k K = Kerf . Hence g = kg and so

f (g ) = f (kg) = f (k)f (g) = eH f (g) = f (g).

Thus ψ is well defined on cosets.

The function ψ is a morphism because

ψ(Kg1Kg2) = ψ(Kg1g2) = f (g1g2) = f (g1)f (g2) = ψ(Kg1)ψ(Kg2).

If ψ(Kg) = eH , then f (g) = eH and g K. Hence the only element in the kernel of ψ is the identity coset K, and ψ is injective. Finally, Im ψ = Imf ,

88 4 QUOTIENT GROUPS

by the definition of ψ. Therefore, ψ is the required isomorphism between G/K and Imf .

Conversely, note that if K is any normal subgroup of G, the map g → Kg is a morphism from G to G/K, whose kernel is precisely K.

By taking f to be the identity morphism from G to itself, the morphism

theorem implies that G/{e} G.

=

The function f : Z → Zn, defined by f (x) = [x], has nZ as its kernel, and

The alternating group An is of index 2 in the symmetric group Sn (by Theorem 3.36) and so is a normal subgroup by Proposition 4.20. It is instructive to obtain this same conclusion from the morphism theorem. If σ is a permutation in Sn, recall that σ is called even or odd according as σ D = D or σ D = −D, where D is the discriminant in n variables (see the discussion leading to Theorem 3.33). Consider the multiplicative group {1, −1}, and define a

tive morphism (verify) and the kernel is the group An of even permutations. Since |Sn| = n!, the morphism theorem and Corollary 4.8 give the following result (and reprove Theorem 3.36).

Example 4.27. Show that the quotient group R/Z, of real numbers modulo 1 is isomorphic to the circle group W = {eiθ C|θ R}.

Solution. The set W consists of points on the circle of complex numbers of unit modulus, and forms a group under multiplication. Define the function f : R → W by f (x) = e2πix . This is a morphism from (R, +) to (W, ·) because

f (x + y) = e2πi(x+y) = e2πix · e2πiy = f (x) · f (y).

This function can be visualized in Figure 4.1 as wrapping the real line around and around the circle.

The morphism f is clearly surjective, and its kernel is {x R|e2πix = 1} = Z.

Therefore, the morphism theorem implies that R/Z = W . The quotient space R/Z is the set of equivalence classes of R under the relation defined by x ≡ y mod Z if and only if the real numbers x and y differ by an integer. This quotient space is called the group of real numbers modulo 1.

90 4 QUOTIENT GROUPS

By Lagrange’s theorem, any proper subgroup must have order 2 or 4. Since any group of order 2 is cyclic, the only proper subgroups that are not cyclic are of order 4 and contain elements of order 1 and 2. There are two such subgroups, K1 = {e, g2, h, g2h} and K2 = {e, g2, gh, g3h}. All the subgroups are illustrated in Figure 4.3.

To find all the quotient groups, we must determine which subgroups are normal.

The trivial group {e} and the whole group D4 are normal subgroups. Since C4, K1, and K2 have index 2 in D4, they are normal by Proposition 4.20, and their quotient groups are cyclic of order 2.

For each of the subgroups above, the left and right cosets containing g are different; therefore, none of the subgroups are normal.

The table above shows that L = {e, g2} is a normal subgroup. The multiplication table for D4/L given in Table 4.4 shows that it is isomorphic to the Klein 4-group.