Group Under Addition or Multiplication?: The set of numbers of the form 3n, where n ∈ Z. Question as in available in description below (photo).

New QMaths 11C Textbook

Would appreciate any help, I'm really finding this challenging.

1 Answer
Mar 4, 2018

#(3ZZ, +)# is a group. #(3ZZ, *)# is not a group.

Explanation:

A group #(S, @)# is a set #S# equipped with an operation #@# with the following properties:

  • #S# is closed under #@# (#a, b in S rarr a@b in S#)
  • #@# is associative (#a, b, c in S rarr (a@b)@c = a@(b@c)#)
  • There is an identity #e in S# (#a in S rarr a@e = e@a = a#)
  • Every element has an inverse (#a in S rarr EE b in S : a@b = b@a = e#)

Note that #(ZZ, +)# is a group, but #(ZZ, *)# is not since it lacks inverse elements.

What about #(3ZZ, +)# ?

  • If #3m, 3n in 3ZZ# then #3m + 3m = 3(m+n) in 3ZZ#. So #3ZZ# is closed under #+#
  • #+# is associative, since it is associative in #ZZ#
  • #0 = 3 * 0 in 3ZZ#. So #3ZZ# contains an identity for #+#
  • If #3m in 3ZZ# then #3(-m) in 3ZZ# and #3m+3(-m) = 0#. So every element has an inverse.

So #(3ZZ, +)# is a group.

What about #(3ZZ, *)# ?

  • If #3m, 3n in 3ZZ# then #(3m) * (3n) = 3(3mn) in 3ZZ#. So #3ZZ# is closed under #*#
  • #*# is associative, since it is associative in #ZZ#
  • #1# is not divisible by #3# so #1 !in 3ZZ#. Hence #3ZZ# contains no identity for #*#
  • #3ZZ# lacks multiplicative inverses. It does not even have an identity,

So #(3ZZ, *)# is not a group.