+738 The group of integers mod 6, \(Z/6Z:=0,1,2,3,4,5\) is a group under the action addition (+). Each of the 6 elements of this group are associated with a unique action, +0, +1, +2, +3, +4, +5.
Every integer is associated with a single number in this set \(Z/6Z\) which is the remainder after dividing by 6. For example, the number 27 is associated with 3 since \(27=4(6)+3\). We say "27 is equivalent to 3 mod 6".
1. Confirm that the set \(Z/6Z:={0,1,2,3,4,5}\) is closed under addition by listing all combinations of pairs of elements and showing that each result is equivalent to one of 0,1,2,3,4, or 5.
2. What is the identity of this group?
3. Find the inverse of each element.
Please help! I don't need the answers to all of these, just whatever you can do, because I don't understand a thing about this! Thank you so much! Any help is greatly appreciated :D
+23246 1) 0 + 0 = 0 1 + 0 = 0 2 + 0 = 0 etc.
0 + 1 = 1 1 + 1 = 2 2 + 1 = 3
0 + 2 = 2 1 + 2 = 3 2 + 2 = 4
0 + 3 = 3 1 + 3 = 4 2 + 3 = 5
0 + 4 = 4 1 + 4 = 5 2 + 4 = 6 = 0
0 + 5 = 5 1 + 5 = 6 = 0 2 + 5 = 7 = 1
2) The identity element is 0 because 0 + n = n + 0 = n for n = 0, 1, 2, 3, 4, 5
3) The inverse of 0 is 0 because 0 + 0 = 0
The inverse of 1 is 5 because 1 + 5 = 5 + 1 = 0
The inverse of 2 is 4 because 2 + 4 = 4 + 2 = 0
The inverse of 3 is 3 because 3 + 3 = 0
+738 thank you so much geno! I'm starting to understand this now!
I just have 1 question, for number 1, why does 1+0=0 and why does 2+0=0? Isn't 1+0=1 and 2+0=2?
+738 and also, this is probably a really dumb question, but the inverse of 4 is 2 and the inverse of 5 is 1 in this situation right? sorry if this was a dumb question, I've just never learned this stuff before.
