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(a) Prove that there exists a unique group homomorphism sgn : S_n → μ₂

Uniqueness For later contradiction, let g : S_n → μ₂ be another group homomorphism not equal to sgn that satisfies the same properties of the sgn function. Then for σ ∈ S_n, we have

but there must exist at least one τ where sgn(τ)≠g(τ). So without loss of generality, let sgn(τ) = 1 and g(τ) = -1. However, then we would have

and

which is not possible with our definition of f_n. Hence we’ve reached a contradiction, and thus g and sgn must be one in the same.

(b)

(c)

We will first prove that the transposition (1,2) has negative sign and then move on to the general case. According to the definition of the sgn function we have that for σ = (1,2),

Now let (i,j) be an abitrary transposition in S_n. Then we have that (i,j) = (1,i)(2,j)(1,2)(1,i)(2,j). This then yields the following

(d)

(e)

(f) Extra Credit

(g) Extra Credit

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(a)

Not the least important of the vectors on the surface of C are e₁,e₂,e₃. Since e₁,e₂,e₃ each have a length of which is the shortest possible length of a vector on the surface of C, a length of one, then their images under L_A must have length of at least one since they must be on the surface. On the other hand, because e₁ + e₂ + e₃ is also on the surface of C then an increase in any one of the sizes of e₁, e₂ or e₃ under L_A will result in L_A(e₁ + e₂ + e₃) being outside of C. Hence L_A will map each of e₁, e₂ and e₃ to one of {±e₁,±e₂,±e₃}. However, because A ∈ GL₃(ℝ), then L_A is an isomorphism and will map e₁, e₂ and e₃ to distinct elements and furthermore to a basis. Therefore L_A preserves the size and angles between e₁, e₂ and e₃, which since those vectors are a basis for ℝ³ makes A orthogonal, and thus G ⊂ O₃.

Now I ∈ G since it preserves C. If A,B ∈ G, then A(C) = C and B(C) = C, implying that (BA)(C) = B(A(C)) = B(C) = C and therefore AB ∈ G. And finally, since A(C) = C then C = A^-1(C), and thus A^-1 ∈ G. These three things, combined with the fact G ⊂ O₃ proven above makes G a subgroup of O₃.

(b)

(c)

π is surjective Let σ ∈ S₃ and define A to be the matrix with i^th column e_σ(i). Therefore according to the previous part of this problem, A ∈ G. Thus Ae_i = e_σ(i) which implies that A(L_i) = L_σ(i) and therefore π(A) = σ.

Ker(π) The kernel of the π will be the set of all matrices A such that A(L_i) = L_i. Therefore A will need to be such that Ae_i = ±e_i. Hence

(d)

We will use the following for vectors to help in our discussion here.

ϕ is surjective Let σ ∈ S₄. We then aim to find a matrix A ∈ G such that Av_i = v_σ(i). So let’s define a linear transformation T : ℝ³ → ℝ³ by its transformation of the basis : T(v_i) = v_σ(i). By equations 2.1 through 2.3 we have that

The definition of T informs us that A ∈ G and that A(d_i) = d_σ(i), thus implying that ϕ(A) = σ.

Ker(ϕ) We know that every element of the kernel will have that d_i = d_i, which implies that e₁ + e₂ + e₃, being contained in a diagonal, will need to be mapped to itself or -e₁ - e₂ - e₃. Hence

(e)

Is det |_G = sgn ∘π This is not true. According to part (c) of this problem, A = is in the kernel of π, meaning that sgn∘π(A) = 1. However, A has a determinant of -1.

Is sgn ∘π = sgn ∘ϕ

Is det |_G = sgn ∘ϕ This is not true. According to part (d) of this problem, A = is contained in the kernel of ϕ, meaning that sgn(ϕ(A)) = 1. However, A has a determinant of -1.

(f) Extra Credit

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(a)

We can think of ϕ as the map that takes s (where D₈ = ⟨s,r⟩) to the symmetry that reflects S across the plane containing e₁ and e₃; i.e. take s to the matrix representation of the linear transformation defined by e₁e₁, e₂- e₂, and e₃e₃. Similarly ϕ will take r to the matrix representation of rotation about e₃ by π∕2.

This high-level definition of ϕ can be made concrete by the formula

for all s^ℓrⁱ ∈ D₈. By noticing that

and

we can see that the following implies that ϕ is a group homomorphism

Now assume that some element s^ℓrⁱ is mapped by ϕ to the identity of G. Then we would have that

The fact that ϕ : D₈ → G is an injective group homomorphism implies that D₈ is isomorphic to a subgroup of G, namely ϕ(D₈).

(b) Extra Credit

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(a) Show that dim_ℂ(V ) = 2 and {1,j} is a ℂ-basis of V

we see that {1,j} spans V .

Not if we were to have 1(a + bi) + j(c + di) = 0, then a + bi + cj -dk = 0 which implies that a = b = c = d = 0. Hence 1 and j are linearly independent.

With these two results, {1,j} is a basis for V , and thus dim_ℂ(V ) = 2.

(b) Show for x ∈ ℍ that “left multiplication by x” is an endomorphism on V .

(c) Show that α : ℍ → End_ℂ(V ) is a ring homomorphism.

(d) Prove that α is injective

(e) Write down the matrix representation of α(i), α(j), and α(k) with respect to the basis {1,j}

(f)

References