Proving the given $\mathbb R^3/H$ $\cong$ $\mathbb R^2$ where $H$ = {$(y,0,0)|y \in \mathbb R$}

Question

So I am given a group $\mathbb R^3$ and a group $H$ = {$(y,0,0)|y \in \mathbb R$}. I have to prove that that $\mathbb R^3/H$ $\cong$ $\mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H \unlhd \mathbb R^3$. So all I know is $\mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: \mathbb R^3 \to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?

Answer 1

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : \mathbb{R}^3 \longrightarrow \mathbb{R}^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = \{(y,0,0) | y \in \mathbb{R} \}$ is its kernel. If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined. Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2).\ \text{Moreover,}\ f(0,0,0) = (0,0)$ The kernel of this map is seen to be all $(x,y,z) \in \mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$. Hence first isomorphism theorem applies and $ \mathbb{R}^3/H \equiv \mathbb{R}^2.$

Answer 2

The first isomorphism theorem asserts that, if $\varphi: A\to B$ is a surjective homomorphism, then $B\cong A/\ker\varphi$. In your problem, you wish to show $\mathbb R^2\cong\mathbb R^3/H $, so a natural guess would be to take $A=\mathbb R^3$ and $B=\mathbb R^2$. Now it remains to construct the homomorphism so that $\ker\varphi=H$. I will leave the rest to you.