Subgroup and index in $\mathbb{Z}^2$

Question

Which of the following are finite index subgroups in $\mathbb{Z}^2$? What is the index??

1. $H=\{(x,y) \in \mathbb{Z}^2 \text{ s.t. } x+y=1\}$

2. $H=\{(x,y) \in \mathbb{Z}^2 \text{ s.t. } x+y=0\}$

3. $H=\{(x,y) \in \mathbb{Z}^2 \text{ s.t. } 2\mid x \}$

4. $H=\{(x,y) \in \mathbb{Z}^2 \text{ s.t. } 2x-2y\equiv 0 \pmod 3 \}$

5. $H=\{(x,y) \in \mathbb{Z}^2 \text{ s.t. } 2x-2y\equiv 0 \pmod 6 \}$

This is what i have done:

1. It cannot be a subgroup under addition operation, so if it is one, it must be under multiplication. So I know it cannot be a subgroup since it doesn't contain $(1,1)$, but I don't know how to prove it properly.

2. This is just a group made from $(x,-x)$. So it has an inverse $(-x,x)$. Let another element be $(b,-b)$, $(b-x, -b+x)$ still belong to $H$, so $H$ is a subgroup. but the index is infinite.

3. It's easy to prove that $H$ is a subgroup, and the index is $2$

4. no idea

5. no idea

Can anyone check if I'm wrong in $1$,$2$ and $3$? If possible, please help me with question $4$ and $5$, I have absolutely no idea.

I really hate it when group theory pop up in number theory...

Answer 1

$\mathbb{Z}^2$ is not a group under multiplication, so I guess that only addition is to be considered.

1. $H$ is not a subgroup, because $(0,0)\notin H$

2. $H$ is indeed a subgroup; why the index is infinite? Hint: consider $f\colon\mathbb{Z}^2\to\mathbb{Z}$ defined by $f(x,y)=x+y$.

3. Consider $f\colon\mathbb{Z}^2\to\mathbb{Z}/2\mathbb{Z}$ defined by $f(x,y)=x+2\mathbb{Z}$.

4. Consider $f\colon\mathbb{Z}^2\to\mathbb{Z}/3\mathbb{Z}$ defined by $f(x,y)=(2x-2y)+3\mathbb{Z}$.

5. The same as with modulo $3$.

---

When one can see a subgroup $H$ of the group $G$ as the kernel of a homomorphism $f\colon G\to G'$, the homomorphism theorem is very handy: it says that $f$ induces an injective homomorphism $$ \bar{f}\colon G/\ker f\to G' $$ which has the same image (some say “range”) of $f$. So we know that the cardinality of $G/\ker f=G/H$ is the same as the cardinality of the image of $f$; but it's also known that the cardinality of $G/H$ is exactly the index of $H$ in $G$.

For instance, in case 3 the homomorphism $f$ is clearly surjective and $$ \ker f=\{(x,y)\in\mathbb{Z}^2:x\in 2\mathbb{Z}\}=H $$ so the index of $H$ is $2$.

Also in cases 2 and 4 the homomorphism is surjective (obvious in case 2, a bit less obvious in case 4) and the kernel is the subgroup you're dealing with. In case 5 the homomorphism is not surjective, but the image is $\{0+6\mathbb{Z},2+6\mathbb{Z},4+6\mathbb{Z}\}$.