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8)
For any n =4, An is simple, i.e., has no proper normal subgroups.
9)
For any n ≥ 3, An is generated by its 3-cycles.
10) Sn can be generated by two elements. In fact, {(1, 2), (1, 2, ..., n)} generates Sn.
(Of course there are subgroups of Sn which cannot be generated by two
elements).
Proof of 4)
The proof presented here uses polynomials in n variables with real
coefficients. Since polynomials will not be introduced until Chapter 3, the student
may skip the proof until after that chapter. Suppose S = {1, ..., n}. If σ is a
permutation on S and p = p(x1, ..., xn) is a polynomial in n variables, define σ(p)
to be the polynomial p(x(1)σ, ..., x(n)σ). Thus if p = x1x2 + x1x3, and σ is the trans-
position (1, 2), then σ(p) = x2x2 + x2x3. Note that if σ1 and σ2 are permutations,
σ2(σ1(p)) = (σ1 ·σ2)(p). Now let p be the product of all (xi -xj) where 1 ≤ i j ≤ n.
(For example, if n =3, p =(x1 - x2)(x1 - x3)(x2 - x3).) If σ is a permutation on S,
then for each 1 ≤ i, j ≤ n with i = j, σ(p) has (xi - xj) or (xj - xi) as a factor. Thus
σ(p) = ±p. A careful examination shows that if σi is a transposition, σi(p) = -p.
Any permutation σ is the product of transpositions, σ = σ1·σ2···σt. Thus if σ(p) =p,
t must be even, and if σ(p) =-p, t must be odd.
Exercise
1 2 3 4 5 6 7
6 5 4 3 1 7 2
1)
Write
as the product of disjoint cycles.
Write (1,5,6,7)(2,3,4)(3,7,1) as the product of disjoint cycles.
Write (3,7,1)(1,5,6,7)(2,3,4) as the product of disjoint cycles.
Which of these permutations are odd and which are even?
2
1
34
Groups
Chapter 2
2)
Suppose (a1, . . . , ak) and (c1, . . . , c ) are disjoint cycles. What is the order of
their product?
3)
Suppose σ ∈ Sn. Show that σ-1(1, 2, 3)σ = ((1)σ, (2)σ, (3)σ). This shows
that conjugation by σ is just a type of relabeling.
Also let τ =(4, 5, 6) and
find τ-1(1, 2, 3, 4, 5)τ.
4)
Show that H = {σ ∈ S6 : (6)σ =6} is a subgroup of S6 and find its right
cosets and its left cosets.
5)
Let A ⊂ R2 be the square with vertices (-1, 1), (1, 1), (1, -1), and (-1, -1),
and G be the collection of all isometries of A onto itself. We know from a
previous exercise that G is a group with eight elements. It follows from Cayley’s
theorem that G is isomorphic to a subgroup of S8. Show that G is isomorphic
to a subgroup of S4.
6)
If G is a multiplicative group, define a new multiplication on the set G by
a ◦ b = b · a. In other words, the new multiplication is the old multiplication
in the opposite order. This defines a new group denoted by Gop, the opposite
group. Show that it has the same identity and the same inverses as G, and
that f : G → Gop defined by f(a) =a-1 is a group isomorphism. Now consider
the special case G = Sn. The convention used in this section is that an element
of Sn is a permutation on {1, 2, . . . , n} with the variable written on the left.
Show that an element of Sn is a permutation on {1, 2, . . . , n} with the variable
written on the right. (Of course, either Sn or Sn may be called the symmetric
group, depending on personal preference or context.)
Product of Groups
The product of groups is usually presented for multiplicative groups. It is pre-
sented here for additive groups because this is the form that occurs in later chapters.
As an exercise, this section should be rewritten using multiplicative notation. The
two theorems below are transparent and easy, but quite useful.
For simplicity we
first consider the product of two groups, although the case of infinite products is only
slightly more difficult.
For background, read the two theorems on page 11.
Theorem
Suppose G1 and G2 are additive groups. Define an addition on G1 × G2
by (a1, a2)+(b1, b2) =(a1 + b1, a2 + b2). This operation makes G1 × G2 into a group.
Its “zero” is (01, 02) and -(a1, a2) =(-a1, -a2). The projections π1 : G1 × G2 → G1
op
op
¯ ¯
Exercise
phic.
Exercise
Suppose G1 and G2 are groups. Show G1 × G2 and G2 × G1 are isomor-
If o(a1) =n and o(a2) =m, find the order of (a1, a2) in G1 × G2.
Chapter 2
Groups
35
and π2 : G1 × G2 → G2 are group homomorphisms. Suppose G is an additive group.
We know there is a bijection from {functions f : G → G1 × G2} to {ordered pairs of
functions (f1, f2) where f1 : G → G1 and f2 : G → G2}. Under this bijection, f is a
group homomorphism iff each of f1 and f2 is a group homomorphism.
Proof
It is transparent that the product of groups is a group, so let’s prove
the last part. Suppose G, G1, and G2 are groups and f = (f1, f2) is a function
from G to G1 × G2. Now f(a + b) = (f1(a + b), f2(a + b)) and f(a) +f(b) =
(f1(a), f2(a))+(f1(b), f2(b)) = (f1(a)+f1(b), f2(a)+f2(b)). An examination of these
two equations shows that f is a group homomorphism iff each of f1 and f2 is a group
homomorphism.
Exercise
Show that if G is any group of order 4, G is isomorphic to Z4 or Z2 ×Z2.
Show Z4 is not isomorphic to Z2 × Z2. Show that Zmn is isomorphic to Zn × Zm iff
(n, m) =1.
Exercise
Suppose G1 and G2 are groups and i1 : G1 → G1 × G2 is defined by
i1(g1) = (g1, 02). Show i1 is an injective group homomorphism and its image is a
normal subgroup of G1 × G2. Usually G1 is identified with its image under i1, so G1
may be considered to be a normal subgroup of G1 × G2. Let π2 : G1 × G2 → G2
be the projection map defined in the Background chapter. Show π2 is a surjective
homomorphism with kernel G1. Therefore (G1 × G2)/G1 ≈ G2.
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