Let $G=\langle \mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$. 
Let $G=\langle\mathbb R \times \mathbb R, +\rangle$ and $H=\{(x,y): y=2x\}$. Show that $H$ is a subgroup of $G$.

I'm trying to show that $H$ is closed and contains inverses (individually). So the ordered pairs have the form $(x,2x)$. Let $(x,y)$ and $(z,w)$ be in $H$. Then
$$\begin{align}
(x,y)+(z,w)&=(x,2x)+(z,2z)\\
&=(x+z,2x+2z)\\
&=(x+z,2(x+z)).
\end{align}$$ Thus $H$ is closed. Now, if $(x,y)=(x,2x)\in H$, then $(-x,-y)=(-x,-2x)$ implies that their sum is $(0,0)$ which is the identity element. Thus $(-x,-y)\in H$ and $H$ is a subgroup of $G$.
Is this correct? Could I also have consider ordered pairs of the form $({1\over 2}y,2x)$ or just consider two of the same elements, $(x,y)$ and done the proof that way? How do I know that I'm setting the problem up right in the first place. I feel like I don't really understand it and I'm just getting lucky sometimes. Any other solutions, hints, advice is greatly appreciated.
 A: Pairs of the form $(\frac 12 y, 2x)$ don't belong to $H$ in general; I think you meant $(\frac y2, y)$. Of course, this could have worked exactly the same way because $y=2x$ is equivalent to $x=\frac y2$.
But more about the proof technique itself: when proving that $H$ is closed under the addition operation, that was fine, but when trying to prove that $H$ is closed under taking inverses, you don't have to calculate what the inverse is if you already know how to compute inverses in the original group (in this case $\mathbb R \times \mathbb R$); you just have to show that the inverse is in $H$. So an equality as simple as
$$
-(x,2x) = (-x, -2x) = (-x, 2(-x)) \in H
$$
would suffice to show that $H$ is closed under taking inverses.
Hope that helps,
A: I will use the one-step subgroup test.
Since $0=2\times 0$, we have $(0,0)\in H$. Hence $H\neq \varnothing$.
Because $H=\{\color{red}{(x,y)}\mid y=2x\}$, we have $H\subseteq G$.
Let $a=(x,y), b=(z, t)\in H$. Then $y=2x$, $t=2z$, and
$$\begin{align}
a-b&=(x-z, y-t)\\
&=(x-z, 2x-2z)\\
&=(\color{green}{x-z}, \color{blue}{2(x-z)}),
\end{align}$$
which is in $H$ since $\color{blue}{2(x-z)}=2(\color{green}{x-z})$; so $a-b\in H$.
Hence $H\le G$.
