ILS 4430 Boston University Math Calculation Worksheet

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Topic 3 Classwork Assignment Problem 1: The set G = {e, a, b, c} under the operation * is an abelian group by the given by the Cayley table. Give examples to show that ⟨G,∗⟩ satisfies Property 1, Property 2, and Property 5 of groups. Property 1: (Closure) For all a, b ∈ G, a ∗ b ∈ G. Property 2: (Associativity) For all a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c (*) This Cayley table is C4 Table 2. Property 5: (Commutivity) For all a, b ∈ G, a ∗ b = b ∗ a. Problem 2: The set G = {e, a, b, c} under the operation * is an abelian group by the given by the Cayley table. Give examples to show that ⟨G,∗⟩ satisfies Property 3 and Property 4 of groups. State clearly the identity element of ⟨G,∗⟩, and the inverse of each element in ⟨G,∗⟩. Property 3: (Identity): For all a ∈ G, there exists an element e ∈ G, called an identity element, such that a ∗ e = e ∗ a = a. (*) This Cayley table is C4 Table 2. Property 4: (Inverse): For all a ∈ G, there exists an element a−1 ∈ G, called the inverse of a such that a ∗ a−1 = a−1 ∗ a = e. Problem 3: Use the given transformation to show that Table 3 and Table 4 for C4 are equivalent. 𝐂𝟒 Table 3: * e a b c e e a b c a a b c e b b c e a c c e a b a→b b→c c→a e→e 𝐂𝟒 Table 4: * e a b c e e a b c a a c e b b b e c a c c b a e Problem 4: C5 = {e, a, b, c, d} is the abstract group that contains five elements. Complete the Cayley table for C5 . * e e e a b c d c a b d c e d e a Topic 3 Homework Assignment Solution Problems: 1,2,3 Problem 1: C6 = {e, a, b, c, d, f} is the abstract group that contains six elements. Construct a Cayley table for C6 . Guess a*a = b. Solution: * e a b c d f e e a b c d f a a b e d f c b b e a f c d c c f d e b a d d c f a e b f f d c b a e Problem 2: C8 = {e, a, b, c, d, f, g, h} is the abstract group that contains eight elements. Construct a Cayley table for C8. Guess a*a = b. Solution: 𝐂𝐂𝟖𝟖 Table C8 = {e, a, b, c, d, f, g, h} * e a b c d f g h e e a b c d f g h a a b c e f g h d b b c e a g h d f c c e a b h d f g d d h g f e c b r f f d h g a e c b g g f d h b a e c h h g f d c b a e Problem 3: C10 = {e, a, b, c, d, f, g, h} is the abstract group that contains ten elements. Construct a Cayley table for C10. Guess a*a = b. Solution: 𝐂𝐂𝟏𝟏𝟏𝟏 Table C10 = {e, a, b, c, d, f, g, h, j, k} * e a b c d f g h j k e e a b c d f g h j k a a b c d e g h j k f b b c d e a h j k f g c c d e a j k f g h d d e a b b k f g h j f f k j h g e d c b a g g f k j a e d c b h h g f k h j b a e d c j j h g f c b a e d k k j h g k d c b a e c f Topic 4 Classwork Assignment D4 is the symmetry group of the 4-gon, or the square. 360° Since Dn = {e, r, r 2 , … , r n−1 , f, rf, r 2 f, … , r n−1 f} with θ = n , D4 = {e, r, r 2 , r 3 , f, rf, r 2 f , r 3 f } with θ= 360° = 4 90°. Problem 1: Label the corners of the square undergoing the given symmetry transformations of D4 . The symmetry transformations of D4 are: e: no transformation r: clockwise rotation of 90° r 2 : clockwise rotation of 180° r 3 : clockwise rotation of 270° f: reflection about symmetry axis rf: clockwise rotation of 90° and then reflection about the symmetry axis 2 r f: clockwise rotation of 180° and then reflection about the symmetry axis 3 r f: clockwise rotation of 270° and then reflection about the symmetry axis Problem 2: Complete the Cayley table for D4 . Use the Cayley table for D4 and the given transformation to show that D4 is equivalent to C8 .∗ 𝐃𝟒 Table D4 = {e, r, r 2 , r 3 , f, rf, r 2 f , r 3 f } * e r r2 r3 f rf 𝐂𝟖 Table C8 = {e, a, b, c, d, f, g, h} r2 f e r r2 r3 r3 f e→e r→a r2 → b r3 → c f→ d rf → f r2f → g r3f → h * e a b c f d rf f r2 f g r3 f h Is D4 abelian? * See Topic 11 homework assignment for C8 . e a b c d f g h Topic 4 Homework Assignment Solution Problems: 1,2 D5 is the symmetry group of the 5-gon, or the pentagon. 360° Since Dn = {e, r, r 2 , … , r n−1 , f, rf, r 2 f, … , r n−1 f} with θ = , D5 = {e, r, r 2 , r 3 ,r 4 , f, rf, r 2 f , r 3 f ,r 4 f } with θ= 360° = 5 72°. n Problem 1: Label the corners of the pentagon undergoing the given symmetry transformations of D5 . Solution: The symmetry transformations of D5 are: e: no transformation r: clockwise rotation of 72° r 2 : clockwise rotation of 144° r 3 : clockwise rotation of 216° r 4 : clockwise rotation of 288° f: reflection about symmetry axis rf: clockwise rotation of 72° and then reflection about the symmetry axis r 2 f: clockwise rotation of 144° and then reflection about the symmetry axis r 3 f: clockwise rotation of 216° and then reflection about the symmetry axis r 4 f: clockwise rotation of 288° and then reflection about the symmetry axis Problem 2: Complete the Cayley table for D5 . Use the Cayley table for D5 and the given transformation to show that D5 is equivalent to C10 .* e→e r→a r2 → b r3 → c r4 → d f→f rf → g r2f → h r3f → j r4f → k Solution: 𝐃𝐃𝟓𝟓 Table D5 = {e, r, r 2 , r 3 , r 4 , f, rf, r 2 f , r 3 f, r 4 f} * e r e e r r r r2 r2 r2 r4 f r4 rf rf r2 f r2 f r4 f r4 f r3 r3 f r3 f r3 f r 2 r2 r3 r 3 r3 r4 4 f r4 f r e e r e r e r r4 f f r3 f r2 r2 r2 f rf r4 f r3 r4 r2 f r3 f Is D5 abelian? No r4 r4 f r3 rf f r3 f r2 f rf r4 f r2 f rf e r f r3 f r2 rf f r4 r3 a b c d f g h j k e e a b c d f g h j k f a a b c d e g h j k f rf b b c d e a h j k f g rf r 2 f c c d e a j k f g h d d e a b b k f g h j f f k j h g e d c b a g g f k j a e d c b h h g f k h j b a e d c j j h g f c b a e d k k j h g k d c b a e r2 f r3 f r4 f f r3 f r4 f r4 f e 3 4 rf r 2 f r 3 f r 4 f r2 f r3 f r4 f f f rf r 2 f r 3 f r4 e r3 r2 r e r2 r3 r r4 r3 r2 r r2 * See Topic 11 homework assignment for C10 . 𝐂𝐂𝟏𝟏𝟏𝟏 Table C10 = {e, a, b, c, d, f, g, h, j, k} * rf r f r f r f 2 r4 r3 e r r4 e c f
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Topic 3Classwork Assignment
Problem 1: The set G = {e, a, b, c} under the operation * is an abelian group by the given by the Cayley
table. Give examples to show that ⟨G,∗⟩ satisfies Property 1, Property 2, and Property 5 of groups.
Property 1:(Closure)
For all a, b ∈ G, a ∗ b ∈ G.
𝑒∗𝑎 =𝑒 ∈𝐺
𝑏∗𝑐 =𝑎 ∈𝐺
𝑐∗𝑎 =𝑏 ∈𝐺
Property 2:(Associativity)
For all a, b, c ∈ G, a ∗ (b ∗ c) = (a ∗ b) ∗ c

(*) This Cayley table is C4 Table 2.

𝑒 ∗ (𝑏 ∗ 𝑐) = 𝑒 ∗ 𝑎 = 𝑎 = 𝑏 ∗ 𝑐 = (𝑒 ∗ 𝑏) ∗ 𝑐
𝑎 ∗ (𝑎 ∗ 𝑏) = 𝑎 ∗ 𝑐 = 𝑏 = 𝑒 ∗ 𝑏 = (𝑎 ∗ 𝑎) ∗ 𝑏
𝑎 ∗ (𝑏 ∗ 𝑐) = 𝑎 ∗ 𝑎 = 𝑒 = 𝑐 ∗ 𝑐 = (𝑎 ∗ 𝑏) ∗ 𝑐
Property 5:(Commutivity)
For all a, b ∈G, a∗ b = b ∗ a.
𝑏∗𝑐 =𝑎 =𝑐∗𝑏
𝑒∗𝑏 =𝑏 =𝑏∗𝑒
𝑎∗𝑐 =𝑏 =𝑐∗𝑎

Problem 2:The set G = {e, ...


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