Mathematics 243, section 3 – Algebraic Structures Problem Set 2 due: September 14, 2012

‘A’ Section

1. Let f : ℤ → ℤ and g : ℤ → ℤ be the indicated functions. In each case, say whether f, g, f ∘ g, g ∘ f are one-to-one (injective) or onto (surjective), both, or neither. Justify your answers.

a. f(x) = 3x, g(x) = 4 − x

b. f(x) = |x|, g(x) = \begin{cases}x & if x is even \\ x − 1 & if x is odd \end{cases} .

2. Consider the binary operation on ℤ given by

x ∗ y = x + y + 5

a. Is ∗ commutative? Why or why not?

b. Is ∗ associative? Why or why not?

c. Is there an identity element for ∗. What is the identity, or why not?

d. Are there any elements of ℤ that have inverses under this operation? What are they, and what are the inverses?

3. Let A = {x, y, z, w} and let ∗ be the binary operation on A given by the following table:

\begin{matrix}∗ & x & y & z & w \\ x & x & y & z & w \\ y & y & y & w & w \\ z & z & w & z & w \\ w & w & w & w & w \end{matrix}

a. Explain how you can tell this operation is commutative.

b. Explain why x is an identity element for ∗.

c. Which elements have inverses and what are the inverses?

d. What is (y ∗ z) ∗ z? Is that the same as y ∗ (z ∗ z)?

‘B’ Section

1. Let f : A → B and g : B → A be mappings. Prove that if f ∘ g is onto and g ∘ f is one-to-one, then f is one-to-one and onto.

2. Let ∗ be an associative binary operation on a set A and assume there is an identity element e for ∗. If a ∈ A has inverses b_{1} and b_{2}, show that b_{1} = b_{2}. Hint: Consider the “product” (b_{1} ∗ a) ∗ b_{2}.

3. Let A be a set and let P(A) be the power set of A as defined in §1 of the text and on Problem Set 1. Let ∗ be the binary operation on P(A) defined by S ∗ T = S ∪ T. Answer the following questions and prove your assertions.

a. Is ∗ associative? Is ∗ commutative?

b. Is there an identity element in P(A) for this operation?

c. What elements of P(A) have inverses for this operation?

d. Make a table like the one in problem 3 of the ‘A’ section for the operation in this problem, when A = {a, b}. List the elements of P(A) in this order on the borders of the table:

∅, {a}, {b}, {a, b}.

Do you notice something?

4. Let × : ℝ^{3} × ℝ^{3} → ℝ^{3} be the vector cross product from multivariable calculus (MATH 241). Recall that this operation is defined by the following formula:

(a_{1}, a_{2}, a_{3}) × (b_{1}, b_{2}, b_{3}) = (a_{2}b_{3} − a_{3}b_{2}, −(a_{1}b_{3} − a_{3}b_{1}), a_{1}b_{2} − a_{2}b_{1}).

a. Show that × is not associative and not commutative.

b. Show that × does satisfy the Jacobi identity:

(a × b) × c = a × (b × c) + b × (c × a)

for all a, b, and c in ℝ^{3}. The + in this formula means the vector sum in ℝ^{3}, defined for vectors d = (d_{1}, d_{2}, d_{3}) and e = (e_{1}, e_{2}, e_{3}) by the rule

d + e = (d_{1}, d_{2}, d_{3}) + (e_{1}, e_{2}, e_{3}) = (d_{1} + e_{1}, d_{2} + e_{2}, d_{3} + e_{3}).

c. Extra Credit In a sense, the additional term b × (c × a) on the right in the Jacobi identity measures the failure of associativity. Using that idea, is a × (b × c) ever equal to (a × b) × c when all three of the vectors are nonzero? Explain. Hint: One way to approach this is to think about the geometric conditions on the three vectors under which it will be true that

b × (c × a) = 0.