Question

1. Which of the following maps define an isomorphism of binary structures? Explain. (a) The function (R, +) → (R, +) given by x → x 2 (b) The function (R>0, ·) → (R>0, ·) given by x → x 2 (c) The function (R>0, +) → (R>0, +) given by x → 2x 2. For each of the following, prove or disprove that it is a group. If it is a group, prove or disprove that it is abelian. (a) (Q +, ·) (b) (R \ {0}, ∗), where a ∗ b = ab 2 3. Assume that φ : (S, ∗) → (S 0 , ∗ 0 ) is an isomorphism of binary structures, and that ∗ is associative. Prove that ∗ 0 is associative. 4. Consider the set H = a −b b a ∈ M2(R) : a 6= 0 or b 6= 0 , with matrix multiplication as the operation. (a) Show that the operation is closed. (b) Define the function φ : H → C ∗ (where C ∗ is the group C \ {0} with multiplication as the operation) by φ : a −b b a 7→ a + bi. Show that φ is an isomorphism of binary structures. (c) Explain briefly why we can now conclude that H is a group. 5. Consider the set V = a 0 0 b ∈ M2(R) : a, b ∈ {1, −1} , with the operation · (matrix multiplication). Note that (V, ·) is a group (you do not need to prove this). Prove that (V, ·) is not isomorphic to U4.

Answer 1

1.(a) it is not an isomorphism since it is not one-one as image of both 1 and -1 is 1.

Note 1 and -1 both belongs to R set of real numbers.

1. (b) it is an isomorphism since it is homomorphism under multiplication

Let a and b are positive real numbers

as f(a.b) = (a.b)^2 = (b.a)^2 as R⁺ is an an Abelian group under multiplication.

= a^2 .b^2 be reversal law.

= f(a). f(b)

so f is homomorphism.

Also f is injection as f(a) = f(b) ⇒ a^2 = b^2 ⇒ a = b as and b are positive

f is also onto since for every b in R^{​​​​​​+}(codomain) there exist real number a in R⁺ such that f(a) = b

i. e. a^2 = b (note : every positive real number has always ca positive square root)

Hence f is an isomorphism

(c) here f is not an isomorphism since f is not a homomorphism

Let a and b are positive real then

f(a + b) = 2(a + b )^2 = 2(a^2 + b^2 + ab + ba)

but f(a) + f(b) = 2a^2 + 2b^2

these two are not equal so f is not homomorphism and so f is not isomorphism.