Subgroups for ordered pairs Question:
$$G=\langle \mathbb{R}\times \mathbb{R},+
\rangle$$
where $H=\{(x,y)\mid y=2x\}$ is a subset of $G$.
So is $H$ a subgroup of the group $G$?
I started of by checking subgroup conditions:
$\Leftarrow$ Let $H=\{(x,y)\mid y=2x\}$ be in $G$ and $B=\{(u,w)\mid w=2u\}$ be in $G$.Thus, $H+B=(x+u,y+w)=(x+u,2x+2u)$ which is in $H+B$
$\Rightarrow$ Let $H=\{(x,y)\mid y=2x\}$ be in $G$ and $H^{-1}=\{(-x,-y)\mid y=2x\}$ be in $G$. Thus $H+H^{-1}=(x\pm x,y\pm y)=(x\pm x,2x\pm 2x)=(0,0)$
Therefore, it is a group. If anyone can verify that would be lovely.
 A: As noted in the comments to your question, you need to show that the set $H$ is a subgroup of the group $G$.  There's a list of things you have to check, and different textbooks have slightly different lists, but they're all equivalent.
Subgroup test (version 1):


*

*Check $H \ne \emptyset$.

*Check that $H$ is closed under the group operation, i.e. $\forall a,b \in H$ we have $ab\in H$.

*Check that $H$ is closed under inverses, i.e. $\forall a \in H$ we have $-a \in H$ (or $a^{-1} \in H$ if our operation is written as multiplication).


Subgroup test (version 2):


*

*Check $H \ne \emptyset$.

*Check that $H$ is closed under composition with inverses, i.e. $\forall a,b \in H$ we have $a-b\in H$ (or $ab^{-1}\in H$ if our operation is written as multiplication).


The version $2$ test is just a shorter, equivalent form of version 1. Some textbooks specify that instead of checking $H \ne \emptyset$ you should check that $e \in H$ (where $e$ is the identity element of $G$), but there's nothing special about choosing $e$; it just gives you a template to follow in your proof.
So your proof in version-1 style might be:


*

*$(x,y) = (0,0)\in H$ since $y=2x$, so $H\ne\emptyset$.

*Let $a=(x_1,y_1), b=(x_2,y_2) \in H$, so $y_1=2x_1$ and $y_2=2x_2$. Then $a+b = (x_1+x_2, y_1+y_2)$, and $y_1+y_2 = 2x_1+2x_2 = 2(x_1+x_2)$ so $a+b \in H$.

*Let $a=(x,y)\in H$, so $y=2x$. Then $-a=(-x,-y)$, and $-y=-2x=2(-x)$ so $-a\in H$.


And in version-2 style:


*

*$(x,y) = (0,0)\in H$ since $y=2x$, so $H\ne\emptyset$.

*Let $a=(x_1,y_1), b=(x_2,y_2) \in H$, so $y_1=2x_1$ and $y_2=2x_2$. Then $a-b = (x_1-x_2, y_1-y_2)$, and $y_1-y_2 = 2x_1-2x_2 = 2(x_1-x_2)$, so $a-b \in H$.

