MATH 433
March 13, 2015
Quiz 7: Solutions
Problem 1. Consider a binary operation ∗ on R given by x ∗ y =
Is (R, ∗) a group? Explain.
p
3
x3 + y 3 for any x, y ∈ R.
Solution: Yes, (R, ∗) is a group.
We need to check four axioms. First of all, the set R is closed under the operation as x∗y = (x3 +y 3 )1/3 ∈
R whenever x, y ∈ R. Further, for any x, y, z ∈ R,
(x ∗ y) ∗ z = (x3 + y 3 )1/3 ∗ z =
(x3 + y 3 )1/3
3
+ z3
1/3
= (x3 + y 3 + z 3 )1/3 ,
3 1/3
x ∗ (y ∗ z) = x ∗ (y 3 + z 3 )1/3 = x3 + (y 3 + z 3 )1/3
= (x3 + y 3 + z 3 )1/3 .
Hence (x ∗ y) ∗ z = x ∗ (y ∗ z) so that the operation is associative.
The equalities x ∗ e = e ∗ x = x are equivalent to (x3 + e3 )1/3 = x. They hold for e = 0 and any x ∈ R.
Therefore 0 is the identity element of the operation ∗.
Given x ∈ R, the equalities x ∗ y = y ∗ x = 0 are equivalent to (x3 + y 3 )1/3 = 0. They hold for y = −x.
Therefore −x is the inverse element of x with respect to the operation ∗.
All four axioms have been checked. Thus (R, ∗) is a group.
Problem 2.
group:
The following is a partially completed Cayley table for a certain commutative
∗
a
b
c
d
a b c d
c
b
c
a
Complete the table.
Solution:
∗ a b c d
a c d a b
b d a b c
c a b c d
d b c d a
Since the group is commutative, the Cayley table is symmetric relative to its diagonal. Also, the
left cancellation property implies that there are no repetitions in each row of the table while the right
cancellation property implies that there are no repetitions in each column. This information is already
sufficient to complete the table. Additional helpful information is that c is the identity element as c2 = c.
Remark. The group in this problem is (Z4 , +). Namely, a stands for [2]4 , b stands for [1]4 , c
stands for [0]4 , and d stands for [3]4 .