# Thread: Quotient (Factor) Groups of Z

1. Okay, I think I was wrong.

2. Originally Posted by mpetnuch
Okay, I think I was wrong.
Here is a way to list all the elements in a nice way. Say $\displaystyle n=3$ instead of using $\displaystyle (x,y) + \left< (3,3)\right>$ as to represent a coset I will just use $\displaystyle (x,y)$.

When $\displaystyle n=3$. Define the "length" of $\displaystyle (a,b)$ to be $\displaystyle a-b$. And as I explained different length give different cosets. So begin with length=0. In that case the cosets have the form $\displaystyle (a,a)$. So it can be $\displaystyle (0,0)$ (identity) or $\displaystyle (1,1)$, $\displaystyle (2,2)$, but not anything else because, say, $\displaystyle (3,3) = (0,0)$ because $\displaystyle 3\equiv 0$. When length=1 we have $\displaystyle (1,0)$, $\displaystyle (2,1)$, and $\displaystyle (3,2)$ and that is it.

So we can make a little table (actually it is a large table because there are infinitely many elements):
length = 0: (0,0), (1,1), (2,2)
length = 1: (1,0), (2,1), (3,2)
length =-1: (0,1), (1,2), (2,3)
...

And we add these elements (cosets) in an obvious way. It should be obvious that if $\displaystyle n=1$ then this table has only a single coloum so it is cyclic but only in that case.

3. ## Proof the groups are isomorphic

Okay! I think I got it for real this time. Your above post helped me realize my mistake, I didn't need the function I was trying to construct to be injective since I could just invoke the first isomorphism theorem.

$\displaystyle f : \mathbb{Z}\times \mathbb{Z}\to \mathbb{Z}_n \times \mathbb{Z}$ by $\displaystyle f(x,y) = (x \bmod n , x - y)$ is an empimorphism.

$\displaystyle f$ is a homomorphism since $\displaystyle f((x,y)+(u,v)) = f((x+u, y+v))$$\displaystyle = (x+u \bmod n, x+u - y -v) = (x \bmod n, x -y) + (u \bmod n, u -v) = f(x,y)+f(u,v)$

$\displaystyle f$ is an epimorphism, since $\displaystyle f$ is a homomorphism and it is a surjective function.

Now the $\displaystyle \ker f = \{ (x,y) : f(x,y) = (0,0) \}= \{ \dots, (-n,-n), (0,0), (n,n), \dots \} = \left<(n,n)\right>$

Hence $\displaystyle \mathbb{Z}\times \mathbb{Z}/ \left< (n, n) \right> \cong \mathbb{Z}_n \times \mathbb{Z}$ by the first isomorphism theorem.

4. Excellent job. You are ready for you exam today.

By the way, the first isomorphism theorem according to your wonderful book is the one mentioned in section 7.1, what you just used is referred to as the fundamental homorphism theorem. However, other algebra books called the homomorphism theorem as the first isomorphism theorem; call your first isomorphism theorem as the second onel; and the books second theorem as the third theorem. So I am just letting you know that different algebra books call it differently, so do not be confused if you ever do that.

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