$G = \Bbb Z \times \Bbb Z$ and $H = \{(a,a): a \in \Bbb Z\}$ what is $G/H$ isomorphic to? $G = \Bbb Z \times \Bbb Z$ and $H = \{(a,a): a \in \Bbb Z\}$ 
I'm trying to find what $G/H$ isomorphic to, but I'm having a really tough time understanding this concept. Can somebody walk me through it? I figure if I have a model, it will be easier to work my way through a few problems...this seemed like the best one to choose as an example.
 A: I'm assuming, based on your tags, that we are treating $G$ as an additive group. In that case, $G$ is the set of 2-tuples with entries in $\mathbb{Z}$, and the group action of addition. $H$ is a subgroup of $G$, where we are restricted to the tuples in which both entries are the same. 
Hence, in $G/H$, anything in the form $(a,a)$ is equal to zero. Hence, if we take an arbitrary element $(x,y) \in G$, then, in $G/H$, we have
$$(x,y) = (x,y)-(y,y) = (x-y,0)$$
because $(y,y) = 0$ in this quotient. Since we can pick $x$,$y$ so that $x-y$ is any integer, we see that this suggests that $G/H \cong \mathbb{Z}$. 
Now, to formally prove this, we can consider the map from $G$ to $\mathbb{Z}$ that takes $(x,y) \to x-y$, and prove that the kernel is equal to $H$.
In general, to determine these quotients, you should try playing around with the relations given by the group you're quotienting over. Then, to make a formal argument, you should construct functions onto the space that you have determined to be the quotient, and show that $H$ is the kernel. 
A: Consider $f:\mathbb{Z}\times\mathbb{Z}\rightarrow \mathbb{Z}$ defined by $f(x,y)=x-y$ show it is surjective and its kernel is $H$.
A: One of the most natural ways to identify a quotient group $G/K$ is to find a homomorphism $\varphi: G \to H$ with $K = \ker(\varphi)$. Put simply, make your subgroup the kernel of some homomorphism. It then follows that $\text{im}(\varphi) \cong G/K$. 
In your case, $G = \mathbb{Z} \times \mathbb{Z}$ and $K = \{(a,a)\,\mid\, a \in \mathbb{Z}\}$. Which function takes a pair $(a,b)$ (an element of $G$) and yields $0$ precisely when $a=b$? An obvious candidate is $\varphi((a,b))=a-b$.  
Finally, just confirm that $\varphi$ is indeed a homomorphism. From where to where? What is its image? 
A: The problem would be easy if $H= \mathbb Z e_1 = \{(a,0): a \in \mathbb Z\}$, right?
Now, $H = \mathbb Z (1,1)$.
So, we try to convert the original problem into this one, by finding a basis of $G$ containing $(1,1)$. This is easy:
$G = \mathbb Z \times \mathbb Z = \mathbb Z f_1 \oplus \mathbb Z f_2$ where $f_1=(1,1)$ and $f_2=(0,1)$.
Then $ H = \mathbb Z f_1 = \mathbb Z f_1 \oplus 0\mathbb Z f_2$.
Therefore, $G/H \cong \mathbb Z /\mathbb Z \times \mathbb Z/0\mathbb Z  \cong \mathbb Z $.
