of the n-gon bisecting the angle between vertices 1 and r + 1. Therefore, gr h always has order 2.

MORPHISMS

Recall that a morphism between two algebraic structures is a function that preserves their operations. For instance, in Example 3.5, each element of the group K of symmetries of the rectangle induces a permutation of the vertices 1, 2, 3, 4. This defines a function f : K → S({1, 2, 3, 4}) with the property that the composition of two symmetries of the rectangle corresponds to the composition of permutations of the set {1, 2, 3, 4}. Since this function preserves the operations, it is a morphism of groups.

Two groups are isomorphic if their structures are essentially the same. For example, the group tables of the cyclic group C4 and ({1, −1, i, −i}, ·) would be identical if we replaced a rotation through nπ/2 by in. We would therefore say that (C4, Ž ) and ({1, −1, i, −1}, ·) are isomorphic.

If (G, ·) and (H, ·) are two groups, the function f : G → H is called a group morphism if

f (a · b) = f (a) · f (b) for all a, b G.

If the groups have different operations, say they are (G, ·) and (H, ), the condition would be written as

f (a · b) = f (a) f (b).

We often use the notation f : (G, ·) → (H, ) for such a morphism. Many authors use homomorphism instead of morphism but we prefer the simpler terminology.

A group isomorphism is a bijective group morphism. If there is an isomor-

phism between the groups (G, ·) and (H, ), we say that (G, ·) and (H, ) are

isomorphic and write (G, ·) (H, ).

=

If G and H are any two groups, the trivial function that maps every element of G to the identity of H is always a morphism. If i: Z → Q is the inclusion map, i is a group morphism from (Z, +) to (Q, +). In fact, if H is a subgroup of G, the inclusion map H → G is always a group morphism.

Let f : Z → {1, −1} be the function defined by f (n) = 1 if n is even, and f (n) = −1 if n is odd. Then it can be verified that f (m + n) = f (m) · f (n) for any m, n Z, so this defines a group morphism f : (Z, +) → ({1, −1}, ·).

Let GL(2, R) be the set of 2 × 2 invertible real matrices. The one-to-one correspondence between the set, L, of invertible linear transformations of the plane and the 2 × 2 coefficient matrices is an isomorphism between the groups

(L, Ž ) and (GL(2, R), ·).

Isomorphic groups share exactly the same properties, and we sometimes identify the groups via the isomorphism and give them the same name. If f : G → H is an isomorphism between finite groups, the group table of H is the same as that of G, when each element g G is replaced by f (g) H .

Besides preserving the operations of a group, the following result shows that morphisms also preserve the identity and inverses.

Proposition 3.19. Let f : G → H be a group morphism, and let eG and eH be the identities of G and H , respectively. Then

(i)f (eG) = eH .

(ii)f (a−1) = f (a)−1 for all a G.

62 3 GROUPS

Hence every cyclic group is isomorphic to either (Z, +) or (Cn, ·) for some n. In the next chapter, we see that another important class of cyclic groups consists of the integers modulo n, (Zn, +). Of course, the theorem above implies that

Any morphism, f : G → H , from a cyclic group G to any group H is determined just by the image of a generator. If g generates G and f (g) = h, it follows from the definition of a morphism that f (gr ) = f (g)r = hr for all r Z.

Proposition 3.21. Corresponding elements under a group isomorphism have the same order.

Therefore, either m and n are both finite and m = n, or m and n are both infinite.

Example 3.22. Is D2 isomorphic to C4 or the Klein 4-group of symmetries of a rectangle?

Solution. Compare the orders of the elements given in Table 3.5. By Proposition 3.21 we see that D2 cannot be isomorphic to C4 but could possibly be isomorphic to the Klein 4-group.

In the Klein 4-group we can write c = a Ž b and we can obtain a bijection, f , from D2 to the Klein 4-group, by defining f (g) = a and f (h) = b. Table 3.6

PERMUTATION GROUPS

A permutation of n elements is a bijective function from the set of the n elements to itself. The permutation groups of two sets, with the same number of elements,

are isomorphic. We denote the permutation group of X = {1, 2, 3, . . . , n} by

(Sn, Ž ) and call it the symmetric group on n elements. Hence Sn = S(Y ) for any n element set Y .

Proposition 3.23. |Sn| = n!

Proof. The order of Sn is the number of bijections from {1, 2, . . . , n} to itself. There are n possible choices for the image of 1 under a bijection. Once the image of 1 has been chosen, there are n − 1 choices for the image of 2. Then

there are n − 2 choices for the image of 3. Continuing in this way, we see that

|Sn| = n(n − 1)(n − 2) · · · 2 · 1 = n! If π : {1, 2, . . . , n} → {1, 2, . . . , n} is a permutation, we denote it by

· · ·

For example, the permutation of {1, 2, 3} that interchanges 1 and 3 is written

which is of order 3! = 6.

If π, ρ Sn are two permutations, their product π Ž ρ is the permutation obtained by applying ρ first and then π . This agrees with our notion of composition of functions because (π Ž ρ)(x) = π(ρ(x)). (However, the reader should be aware that some authors use the opposite convention in which π is applied first and

S3, calculate π Ž ρ and ρ Ž π .

64 3 GROUPS

Under ρ, 1 is mapped to 3, and under π , 3 is mapped to 2; thus under π Ž ρ, 1 is mapped to 2. Tracing the images of 2 and 3, we see that

and π(x) = x if x / {a1, a2, . . . , ar }, is called a cycle of length r or an r-cycle.

Since π r fixes all the other elements, it is the identity permutation. But none of the permutations π, π 2, . . . , π r−1 equal the identity permutation because they

We can either calculate a product of cycles from the permutation representation or we can use the cycle representation directly. Let us calculate π Ž ρ Ž σ from their cycles. Remember that a cycle in S4 is a bijection from {1, 2, 3, 4} to itself, and a product of cycles is a composition of functions. In calculating such a composition, we begin at the right and work our way left. Consider the effect of π Ž ρ Ž σ on each of the elements 1, 2, 3, and 4.

Permutations that are not cycles can be split up into two or more cycles as follows. If π is a permutation in Sn and a {1, 2, 3, . . . , n}, the orbit of a under π consists of the distinct elements a, π(a), π 2(a), π 3(a), . . .. We can split a permutation up into its different orbits, and each orbit will give rise to a cycle.

π(1) = 3, π 2(1) = π(3) = 8, π 3(1) = 4, and π 4(1) = 1; thus the orbit of 1 is {1, 3, 8, 4}. This is also the orbit of 3, 4, and 8. This orbit gives rise to the cycle

is written as a product of disjoint cycles, it does not matter in which order

Proposition 3.27. Every permutation can be written as a product of disjoint cycles.

Proof. Let π be a permutation and let γ1, . . . , γk be the cycles obtained as described above from the orbits of π . Let a1 be any number in the domain of π , and let π(a1) = a2. If γi is the cycle containing a1, we can write γi = (a1a2 · · · ar ); the other cycles will not contain any of the elements a1, a2, . . . , ar

π = γ1 Ž γ2 Ž · · · Ž γk , because they both have the same effect on all the numbers in the domain of π .

Corollary 3.28. The order of a permutation is the least common multiple of the lengths of its disjoint cycles.

Proof. If π is written in terms of disjoint cycles as γ1 Ž γ2 Ž · · · Ž γk , the order of the cycles can be changed because they are disjoint. Therefore, for any integer