Question

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

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

Answer 1

`(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.