Consider the dire t pro du t Z × V where Z is the group of integers (under ad-dition) and V is the Klein 4-group with elements {e, a, b, c} and multipli ation∗ dened by the table b elow:* e a b ce e a b ca a e c bb b c e ac c b a e(a) Find the pro du t (5, c)(2, a). [1 mark℄(b) Find the inverse of (−3, b). [1 mark℄( ) Find the order |Z × V |. [1 mark℄(d) Is Z × V ab elian? (Justify your answer.) [1 mark℄(e) Find the order of (0, a) and the order of (1, b)

(a) The product (5, c)(2, a) in Z × V is computed component-wise. For the first component, we add the integers, and for the second component, we use the multiplication table of V. So, (5, c)(2, a) = (5+2, c*a) = (7, e) according to the multiplication table.

(b) The inverse of an element in a direct product is also computed component-wise. The inverse of an integer under addition is its negation, and the inverse of an element in V is itself (since V is a group of order 2). So, the inverse of (−3, b) is (-(-3), b) = (3, b).

(c) The order |Z × V| of the direct product is the product of the orders of the individual groups. The order of Z (the group of integers under addition) is infinite, and the order of V (the Klein 4-group) is 4. So, |Z × V| = infinity * 4 = infinity.

(d) Z × V is abelian if and only if both Z and V are abelian. The group of integers under addition is abelian because addition of integers is commutative. The Klein 4-group is also abelian, as can be seen from its multiplication table. Therefore, Z × V is abelian.

(e) The order of an element in a direct product is the least common multiple of the orders of its components. The order of 0 in Z is infinite, and the order of a in V is 2 (since aa = e). So, the order of (0, a) is lcm(infinity, 2) = infinity. The order of 1 in Z is infinite, and the order of b in V is 2 (since bb = e). So, the order of (1, b) is lcm(infinity, 2) = infinity.