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## Homework Statement

Show that Z18/M isomorphic to Z6 where m is the cyclic subgroup <6>

operation is addition

## The Attempt at a Solution

M = <6> , so M = {6, 12, 0}

I figured I could show that Z18/M has 6 distinct right cosets if I wanted to do M + 0 = {6, 12, 0} M + 1 = {7, 13, 1} M + 2 = {8, 14, 2} etc and after M + 5 , they would start to repeat

After that I do not know how you show isomorphism. I know it has to satisfy three conditions, injective, surjective, and homomorphism, but I do not know what the function is even, so I'm assuming it is f(a) = M + a for a an element of Z18

so injective : if f(a) = f(b) so ( M +a = M + b ), then a = b.

M + a = {6 +a, 12 + a, a} and M + b = {6 + b, 12 + b, b}

then 6 + a = 6 + b

12 + a = 12 + b

a = b , so I guess I need to show that each of those three equations show a = b, and I will use cancellation of 6, and 12 and we get a = b, so f is injective (but I don't understand how it is injective

surjective: let a be arbitrary element of Z18, then f(a) = M + a = {6 + a, 13 + a, a} so a would be in Z18/M , so it is surjective

homomorphism: f(a + b) = f(a) + f(b) , so

M + ( a + b) = M + a + M + b = f(a) + f(b) this result came from definition of a quotient group from my book which said that G/N (N a normal subgroup) is a group under the operation defined by(Na)(Nc) = N(ac)

Now my book gave a hint saying show that 1 +M generates all of Z18/M, and so Z18/M is cyclic and then show that the order of (1 + M) is 6, however, I do not quite understand how you would show that a set has order since 1 + M = {7, 13, 1} right?

much less showing that 1 + M generates all of Z18/M. thanks