Question

5. Determine if the following sets along with the given operations form groups. If so, determine the identity element and whether or not they are Abelian. If not, explain why.

(a) GL(n, Z) where ∗ is matrix multiplication. This is the collection of all n × n nonsingular matrices with integral entries.

(b) Sym(X) where X is a nonempty set and f ∈ Sym(X) if and only if f : X → X is bijective where ∗ is composition.

(c) Aff(1, R), where Aff(1, R) := {fa,b : R → R : fa,b(x) = ax + b, a, b ∈ R, a 6= 0} and ∗ is composition. These are called the one-dimensional affine functions. What happens if we allow a = 0?

(d) T := {z ∈ C : |z| = 1} where ∗ is complex multiplication. We will again encounter T in later sections.

(e) SL(2, Z) where A ∈ SL(2, Z) if and only if A is a 2 × 2 matrix of integers for which det A = 1. What about SL(n, Z) where n ∈ N?

Answer 1

a) with operation is not a group. Take, . Let . Now is inverse of A, which do not belong to . So is not a group.

is identity in .

b) with composition as multiplication.

, then are bijective. Then we can define such that and . is identity map from to .

, then . Composition of functions is associative. So, . is injective, that means . So is injective.

Now let , is surjective, so there exists , such that . Now is also surjective. So there exists , such that . Then . So again surjective, in other words it is bijective. So, .

is identity element in . But it may not be abelian. For example, take symmetric group of 3 elements, , which is non-abelian.

c) .

Let .

. Of course, and . So .

.

Now if . Put , given .

So is inverse of .

Function composition is associative. So is a group.

.

Now if we choose , we will have . Take . So it is non-abelian.

Of course, if we allow we can not find inverse of .

d)

and . Then .

Then

. As and . So .

, So . So .

, which is identity.

Associativity and commutativity comes from asscociativity and commutativity of .