Proving that $\mathbb{R}^4 / H \cong \mathbb{R}^2$

Question

⁸Let $H = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 - 2x_2 = 0 = x_3 + x_4 \}$. Group operation for $\mathbb{R}^4$ and $\mathbb{R}^2$ is standard vector addition. I need to prove that $\mathbb{R}^4 / H \cong \mathbb{R}^2$.

So far I managed to prove that $H$ is a subgroup of $\mathbb{R}^4$ and since $\mathbb{R}^4$ is obviously an abelian group, then $H$ is a normal subgroup, hence we can create factor group $\mathbb{R}^4 / H$. Then, my observation is that $$ H = \{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = 2 \cdot x_2 \land x_3 = -x_4 \} = \{ (2x_1, x_1, x_2, -x_2) \in \mathbb{R}^4 : x_1, x_2 \in \mathbb{R} \} $$ so intuitively I feel like there must be isomorphism between $\mathbb{R}^4 / H$ and $\mathbb{R}^2$.

My intuition tells me that I have to use First Isomorphism Theorem for groups. If my intuition is good, then I have to find some homomorphism $f$ such that $\ker f = H$, and then I could plug $f$ into the First Isomorphism Theorem to get $\mathbb{R}^4/H \cong \mathbb{R}^2$. This means $f$ has to be some function $f : G \to \mathbb{R}^2$, but I don't know which $G$ I should pick.

Any hints on how to procced my proof would be much appreciated, thank you in advance!

Answer 1

Define$$\begin{array}{rccc}f\colon&\Bbb R^4&\longrightarrow&\Bbb R^2\\&(x_1,x_2,x_3,x_4)&\mapsto&(x_1-2x_2,x_3+x_4).\end{array}$$Then $f$ is surjective and $\ker f=H$. Therefore, $f$ induces an isomorphism from $\Bbb R^4/H$ onto $\Bbb R^2$.

Answer 2

Note that $\mathbb R^4/H$ is a vector space and has dimension $2$. So suppose that $\{e_1'+H, e_2'+H\}$ is a basis of $\mathbb R^4/H$. (Consider $e_1'=(0,0,1,1), e_2'=(-1,2,0,0)$).

$\mathbb R^2$ is also of dimension $2$. Suppose that $\{e_1,e_2\}$ is a basis of $\mathbb R^2$.

Now consider the linear transformation from $\mathbb R^4/H$ to $\mathbb R^2$ defined as : $e_1'+H\mapsto e_1, e_2'+H\mapsto e_2$. Note that this linear transformation is an isomorphism.