Quotient Group of $\mathbb R\times\mathbb R$ proofs

Question

In each of the following $H$ is a subset of $\mathbb R\times\mathbb R$. I want to:

a) Prove that $H$ is a normal subgroup of $\mathbb R\times\mathbb R$.

b) In geometrical terms, describe he elements of the quotient group.

c) Describe the operation of $G/H$.

1. $H=\{(x,0): x\in \mathbb R\}$
2. $H=\{(x,y): y=-x\}$
3. $H=\{(x,y): y=2x\}$

For 1. I believe that $\mathbb R\times\mathbb R$ is abelian , thus making $H$ normal as long as it is a subgroup of $G$. So suppose some $x_1, x_2 \in H$, then $x_1+(x_2)=(x_1, 0)+(x_2,0)=(x_1+x_2, 0) \in H$. $H$ so its closed under addition. Now I need to prove negation. Geometrically, $H$ is all the lines parallel to the horizontal line through the x-axis. I think the operation would be addition.

For 2. Similar to 1, showing $H$ is a subgroup. Let some $y_1, y_2 \in H$ then $y_1+y_2=(-x_1)+(-x_2)=-(x_1+x_2) \in H$. So H is closed. Now to prove negation, that is: $(-x,-y) \rightarrow -y=-(-x) \rightarrow -y=x$ which is the negation of $y=-x$. Geometrically, $H$ is all the lines parallel to the line through the origin with a negative slope. I believe the operation is addition.

For 3. Proving a subgroup: suppose some $y_1,y_2, x_1,x_2 \in H$ then $y_1+y_2=2x_1+(2x_2)=2(x_1+x_2) \in H$. For negation: Geometrically, $H$ is all the lines parallel to the line through the origin with slope 2. I believe the operation is addition.

I have edited to where I am currently at. I am working through negations.

Answer 1

My thoughts are, without the operation being specified, we must be considering $\mathbb{R}^2$ under addition. I don't think there are any other operations you could assume would be on $\mathbb{R}^2$, without it being explicitly provided. Unfortunately, this makes the question a little simple and repetitive, but I'll do part 2, and you can fill in the rest.

(a) First, note that $(0, 0)$, the group identity, is in $H$, as $0 = -0$.

Suppose $(x_1, y_1), (x_2, y_2) \in H$. Then $y_1 = -x_1$ and $y_2 = -x_2$, which implies $$y_1 + y_2 = -(x_1 + x_2) \implies (x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2) \in H.$$ Thus $H$ is closed under the group operation. Note also that given $(x, y) \in H$, the inverse element $(-x, -y)$ satisfies $-y = -(-x)$, since $y = -x$, so $(-x, -y) \in H$. Therefore $H$ is a subgroup. (As you said, it's normal because $\mathbb{R}^2$ is abelian.)

(b) Two points $(x_1, y_1), (x_2, y_2)$ are in the same coset if and only if $$(x_1, y_1) - (x_2, y_2) = (x_1 - x_2, y_1 - y_2) \in H.$$ Therefore, we have $y_1 - y_2 = -(x_1 - x_2)$, or equivalently, $x_1 + y_1 = x_2 + y_2$. So really, we just need to look at the sum of the coordinates to determine which coset we're in - when the sums are the same, the points are in the same coset, and when they're different, they're in different cosets. Our cosets will all be the solution sets of equations of the form $x + y = k$, for fixed $k$.

Geometrically, these equations describe all the parallel lines to $y = -x$. Note that, as cosets should, these partition the plane.

(c) Now that we have a parameter to describe which coset $(x, y) \in \mathbb{R}^2$ belongs to (i.e. $x + y$), we can describe the operation on $G/H$. If we add cosets $(x_1, y_1) + H$ and $(x_2, y_2) + H$, then the resulting coset is $(x_1 + x_2, y_1 + y_2) + H$, which has the parameter $(x_1 + y_1) + (x_2 + y_2)$.

That is, the operation on $G/H$ takes the sets of all pairs whose coordinates sum to $k_1$ and $k_2$ respectively, and produces the set of pairs whose coordinates sum to $k_1 + k_2$.