Jon P. answered • 08/29/15
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This is a group closed under multiplication.
If you multiply any member of the set by any other member, the result will still be a member of the set.
The other requirements for a set to be a group are also met:
- There is an identity element for multiplication, namely 1
- Each element has an inverse. (1 x 1 = 1, i x -i = -i x i = 1, -i x i = i x -i = 1, and -1 x -1 = 1)
- Without working out all of the possibilities, the associate property also exists under multiplication.
Jon P.
tutor
Well yes, it is closed under division. However, your original question referred to the set as a group, but the set is NOT a group under division.
A group has several requirements in addition to closure, including that there must be an identity element, usually referred to as "e". For every member a, e · a = a · e = a.
This is true when the operation is multiplication, in which case 1 is the identity. But when the operation is division, it is not true. For example, i / 1 = i. But 1 / i = -i ≠ i. So 1 does not qualify as the identity under division. And if you try every other element as a possible identity element, you'll see that none of them would qualify.
So using division as the operation, the set is not a group.
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08/29/15
JOHN F.
08/29/15