Quick viewing(Text Mode)

Math 444 - Homework 1 Solutions

Math 444 - Homework 1 Solutions

Math 444 - Homework 1 Solutions

1. Show that if G is a , then Sym(G) - the set of of G - is also a group.

Proof. Let G be a group. First, we show that composition is a binary on Sym(G), i.e. that this operation is closed. Suppose that φ and ψ are automorphisms of G, i.e. that they are bijective . Since the composition of two is again a , φ ψ is a bijection as well. Similarly, the composition of two homomorphisms is a : if g , g ◦ G, then 1 2 ∈ (φ ψ)(g g ) = φ(ψ(g g )) since function composition is associative ◦ 1 2 1 2 = φ(ψ(g1)ψ(g2)) since ψ is a homomorphism

= φ(ψ(g1))φ(ψ(g2)) since φ is a homomorphism = (φ ψ)(g )(φ ψ)(g ) since function composition is associative ◦ 1 ◦ 2 and thus φ ψ is an . We are now ready to prove the remaining criteria: ◦ Associativity: Function composition is always associative. Identity: The automorphism id : g g is the since for any φ Sym(G), we have �→ ∈ (id φ)(g) = id(φ(g)) = φ(g) ◦ for all g G. In other words, id φ = φ. ∈ ◦ 1 Inverses: Let φ Sym(G). Since φ is a bijection, it has an inverse φ− , and this is also a homomorphism. ∈To see this, let g , g G and let h , h G be such that φ(h ) = g and 1 2 ∈ 1 2 ∈ 1 1 φ(h2) = g2 - we know these exist since φ is a bijection. Since φ is a homomorphism, we have the following of equalities:

1 1 φ− (g1g2) = φ− (φ(h1)φ(h2)) 1 = φ− (φ(h1h2))

= h1h2 1 1 = φ− (g1)φ− (g2)

1 Thus, φ− is a homomorphism as well and we’re done. Hence, Sym(G) is a group. 2. The O(2) is the set

T T A M2 2(R) A A = AA = I { ∈ × | } with the of multiplication, where M2 2(R) refers to the set of 2 2 matrices with real entries. × × (a) Verify that O(2) is a group. (b) For each A O(2) and each vector v R2, define A v to be the usual matrix-vector . ∈ ∈ · Verify that this defines a O(2) ↷ R2. (c) The special orthogonal group SO(2) is the subgroup of O(2) which consists of all elements with determinant 1. i. Show that SO(2) is a subgroup of O(2). ii. Show that SO(2) is isomorphic to the group of unit complex numbers under multiplication. (Hint: recall that each unit complex number can be written as eiθ for some θ)

(d) Recall that the D4 is the group of symmetries (rotations and reflections) of a square. Find a subgroup of O(2) which is isomorphic to D4 and explain why.

(e) Find a subgroup of O(2) which is isomorphic to the Sym3 and explain why.

(a) Proof. We begin by checking that is a binary operation on O(2). Let A and B be matrices in O(2), i.e. suppose that AT A = AAT = I and BT B = BBT = I. Then (AB)T (AB) = BT AT AB = BT IB = BT B = I and (AB)(AB)T = ABBT AT = AIAT = AAT = I, so AB O(2) as well. We can now check the remaining criteria. ∈ Associativity: Matrix multiplication is always associative. Identity: The identity matrix I satisfies IT I = IIT = I, so it is an element of O(2). Fur- thermore, we know that IA = AI = A for any other 2 2 matrix A, so I is the identity of O(2). × Inverses: By definition of the property AAT = AT A = I, we can see that AT is the inverse of A, and thus every element of O(2) has an inverse - its transpose. All that remains is to check that AT O(2), which is easily verified: ∈ (AT )T AT = AAT = I and AT (AT )T = AT A = I so AT O(2). ∈ Thus, O(2) is a group. (b) Proof. By definition of the matrix-vector product, we know that for any vector v R2, Iv = v and A(Bv) = (AB)v for any 2 2 matrices A and B, so this defines a group action. ∈ × (c) i. Proof. Since the identity matrix has determinant 1, we know SO(2) is nonempty. It then suffices 1 to show that if A, B SO(2), then A− B is in SO(2) as well. Since O(2) is a group, we already 1 ∈ 1 know that A B O(2) and all that remains is to show that det(A− B) = 1. Indeed, ∈ 1 1 det(A− B) = det(A− ) det(B) 1 = det(B) det(A) = 1

1 and thus A− B SO(2), so SO(2) is a subgroup. ∈

Page 2 ii. Proof. Let U(1) denote the group of unit complex numbers under multiplication1. Then U(1) consists of elements of the form eiθ, where θ is a real number. Define the ψ : U(1) SO(2) → by sending eiθ to cos θ sin θ , sin θ −cos θ � � the counter-clockwise matrix through an angle of θ. Despite the fact that each unit complex number may be represented as both eiθ and eiφ for distinct real numbers θ and φ, the map ψ is well-defined since eiθ = eiφ if and only if θ and φ differ by an integer multiple of 2π, which is precisely when cos θ = cos φ and sin θ = sin φ. This is a homomorphism by the following equalities: cos θ sin θ cos φ sin φ ψ(eiθ)ψ(eiφ) = sin θ −cos θ sin φ −cos φ � � � � cos θ cos φ sin θ sin φ cos θ sin φ sin θ cos φ = sin θ cos φ +−cos θ sin φ − sin θ sin φ +−cos θ cos φ � − � cos(θ + φ) sin(θ + φ) = sin(θ + φ) −cos(θ + φ) � � = ψ(ei(θ+φ)) Furthermore, this map is injective since it has trivial kernel: if ψ(eiθ) is the identity in SO(2), then cos θ = 1 and sin θ = 0, so θ is an integer multiple of 2π and thus eiθ = 1.

To complete the proof, we need only show that ψ is sujective, i.e. each element of SO(2) is a of the form given above. To do so, we will consider what each element does T T to the standard basis e1 = (1, 0) and e2 = (0, 1) . Recall that the dot product u v can be written using matrix multiplication as uT v. For any A SO(2), we then have the fo·llowing: ∈ u v = uT v · = uT AT Av = (Au)T (Av) = (Au) (Av) · Thus, e1 e1 = (Ae1) (Ae1) and e2 e2 = (Ae2) (Ae2), so the linear transformation represented by A send· s the standa· rd basis to tw·o unit vecto·rs. Also, this transformation preserves the angle between the standard basis vectors since Ae1 Ae2 = e1 e2. Since Ae1 and Ae2 are the columns of A with respect to the standard basis, this ·tells us th·at if a c A = , b d � � then a2 + b2 = 1, c2 + d2 = 1, and ac + bd = 0. Since det A = 1, we also know that ad bc = 1. By multiplying both sides of ac + bd = 0 by c and ad bc = 1 by d, we have − − ac2 + bcd = 0 and ad2 bcd = 1 − and by adding, we have a(c2 + d2) = d, i.e. a = d. Plugging in, we find b = c and thus A is of the form − a b A = . b −a � � Since a2 + b2 = 1, we know that a = cos θ and b = sin θ for some real number θ, and we’re done. 1This notation comes from the fact that this group is called the unitary group U(n) for n = 1.

Page 3 (d) Proof. To find a subgroup of O(2) which is isomorphic to D4, we consider the linear transformations R2 R2 which send a specified square to itself - for example, the square with endpoints ( 1, 0), (0, →1). As described in the part (c), the counter-clockwise rotation of the plane through an±angle of π±/2 is given by the matrix 0 1 A = 1 −0 � � and this certainly sends the square to itself. In fact, we can see that the induced symmetry on the square is the rotation r in our usual description of D4. Additionally, we know that A is an element of O(2) since it lies in SO(2) by part (c). Similarly, the reflection of R2 which preserves the line x = y is given by the matrix 0 1 B = 1 0 � � and this transformation preserves the square, inducing the symmetry s. Since the transpose of B is clearly its inverse, we know B O(2). ∈ We then claim that the set I, A, A2, A3, B, BA, BA2, BA3 { } forms a subgroup of O(2) which is isomorphic to D4. One way to check this is to compute the multiplication tables for each group and verify that they are equivalent; another is to check that the map sending r A and s B respects the necessary relations between r and s. In other words, if we can sho�→w that A4 =�→I, B2 = I, and AB = B3A, then this implies that the map is an . Indeed, we have the following:

4 0 1 A4 = = I 1 −0 � � 2 0 1 B2 = = I 1 0 � � 0 1 0 1 1 0 AB = = 1 −0 1 0 −0 1 � � � � � � 3 0 1 0 1 0 1 0 1 1 0 BA3 = = = 1 0 1 −0 1 0 1 0 −0 1 � � � � � � �− � � � Thus, the relations hold and our map is an isomorphism2.

(e) Proof. Recall that the symmetric group Sym3 is the of the triangle, i.e. that Sym3 is isomorphic to the dihedral group D3. So we may complete this problem in essentially the same way as part (d), but by studying a triangle instead. This time we consider the triangle with vertices at the points (1, 0), ( 1/2, √3/2), ( 1/2, √3/2) and define the rotation matrix − − − cos 2π/3 sin 2π/3 1/2 √3/2 A = − = − − sin 2π/3 cos 2π/3 √3/2 1/2 � � � − �

which lies in SO(2) and thus O(2), and induces the element r D3. Similarly, define B to be the matrix which represents the vertical reflection fixing the x-axis∈:

1 0 B = 0 1 � − � 2This technique for verifying may be unfamiliar to you, in which case a multiplication table or a Compelling Definition will suffice instead. In later chapters, we will make precise what we mean by “relations” here.

Page 4 3 This also lies in O(2) and induces the element s D3. As in part (d), we check the relations A = I, B2 = I, and AB = B2A: ∈ 3 1/2 √3/2 A3 = − − = I √3/2 1/2 � − � 2 1 0 B2 = = I 0 1 � − � 1/2 √3/2 1 0 1/2 √3/2 AB = − − = − √3/2 1/2 0 1 √3/2 1/2 � − � � − � � � 2 1 0 1/2 √3/2 1 0 1/2 √3/2 1/2 √3/2 BA2 = − − = − = − 0 1 √3/2 1/2 0 1 √3/2 1/2 √3/2 1/2 � − � � − � � − � �− − � � � Thus, our map satisfies the requisite relations and thus the set

I, A, A2, B, BA, BA2 { } forms a subgroup of O(2) which is isomorphic to D3.

Page 5

Top View