# Thread: what is [x]?

1. ## what is [x]?

I have to prove Y: Z -> Zm defined by Y(x) = [x] is a homomorphism, but I don't remember what [x] means. Any help would be greatly appreciated! Thanks!

2. Ok, I figured out [x] (it's the congruence class of x, yay!), but I'm given that the mapping in my first post is a homomorphism and I'm told to prove it. In order to prove that, I would have to show [x + y] = [x] + [y].
I don't see how that is true; for mod 3, [0] = {..., -6, -3, 0, 3, 6, ...} and
[1] = {..., -5, -2, 1, 4, 7, ...}. According to that equation, [0 + 1] = [0] + [1] = {..., -11, -5, 1, 7, 13, ...} which would be [1] with respect to mod 6.
Am I on the right track here, or am I completely off?

3. Recall the definition of homomorphism.

You need to show that Y(xy)=Y(x)Y(y) for all x,y in Z.

So, Y(xy)=[xy] =[x][y]=Y(x)Y(y). Since you can always choose x to be the identity or 1, so [xy]=[x][y] will always be true if x,y in Z.

4. Originally Posted by mdoyle
Ok, I figured out [x] (it's the congruence class of x, yay!), but I'm given that the mapping in my first post is a homomorphism and I'm told to prove it. In order to prove that, I would have to show [x + y] = [x] + [y].
I don't see how that is true; for mod 3, [0] = {..., -6, -3, 0, 3, 6, ...} and
[1] = {..., -5, -2, 1, 4, 7, ...}. According to that equation, [0 + 1] = [0] + [1] = {..., -11, -5, 1, 7, 13, ...} which would be [1] with respect to mod 6.
Am I on the right track here, or am I completely off?
You're on the right track but you're not adding the cosets properly. Remember that if $A, B \subset G$ then $A+B = \{a+b : a \in A, b \in B\}$. For instance $3+7=10$ should be in there.