Understanding the quotient of infinite groups $\mathbb{R}^2/H$ where $H = \{(a, 0): a\in \mathbb{R}\}$

Question

Define $H = \{(a, 0): a\in \mathbb{R}\}$.

Without using the fundamental homomorphism theorem, how would we know what $\mathbb{R}^2/H$ is? The quotient group is $\{H, (x, y) + H, (x_2, y_2) + H, \dots \}$. Intuitively, each coset in the quotient group is a horizontal line crossing through the point $(x_i + a, y_i)$ or just $(0, y_i)$. And based on other material I've covered, I intuitively think $\mathbb{R}^2/H \cong \mathbb{R}$, but I wouldn't know how to show it.

Answer 1

We'll do this two ways: in words, and then with an isomorphism.

Let's show that every element in the quotient group can be written like $H + (0,y)$.

So start with some element $(x,y)+H$ in the quotient group. What this means is that for any element $h \in H$, our element $(x,y)+H$ is the same element as $ (x,y) + h + H$. But $(-x,0)$ is in $H$. And so our element $(x,y) + H = (0,y) + H$. In this way, it's easy to see that for any $y \in \mathbb{R}$, there is an element in $\mathbb{R}^2/H$ that looks like $(0,y) + H$.

Further, no two elements $(0,a) + H$ and $(0,b) + H$ are the same when $a \neq b$, as $H$ does not affect the second coordinate. The typical group operations of $\mathbb{R}$ still work on the second coordinate as well, for the same reason that $H$ does not affect the second coordinate.

Thus $\mathbb{R}^2 / H \cong \mathbb{R}$.

Alternately, what we have considered is this map:

$$\begin{align}\mathbb{R}^2 / H &\to \mathbb{R}\\ (x,y) + H &\mapsto y\end{align} $$

The first notable paragraph above shows it's surjective. The second shows that it's well-defined and injective. I sort of wave my hands at showing it's a homomorphism, but it is (and you can check it!). Thus we have explicitly given an isomorphism.

Answer 2

The most straightforward way to show it is to write down the isomorphism that you’ve discovered and then prove that it is an isomorphism. Your map is

$$h:\Bbb R^2/H\to\Bbb R:\langle x,y\rangle+H\mapsto y\;.$$

You need only show that $h$ is a well-defined bijection with the homomorphism property.

A small notational point: $H$ has uncountably many cosets, so you can’t actually index them by the natural numbers.

Answer 3

An element of $\mathbb{R}^2/H$ is, as you pointed out, of the form $(x, y) + H$ for some $(x, y) \in \mathbb{R}^2$. As $H = \{(a, 0) \mid a \in \mathbb{R}\}$, we have $(x, 0) + H = (0, 0) + H = H$, so $$(x, y) + H = (0, y) + (x, 0) + H = (0, y) + H.$$ So every element of $\mathbb{R}^2/H$ is of the form $(0, y) + H$ for some $y$. When are two such elements the same? Suppose $(0, y_1) + H = (0, y_2) + H$, then $$(0, y_1) - (0, y_2) + H = H$$ so $(0, y_1-y_2)+H = H$. Now $(a, b) + H = H$ if and only if $(a, b) \in H$ so $(0, y_1 - y_2) \in H$. That is, $(0, y_1-y_2) = (a, 0)$ for some $a \in \mathbb{R}$; in particular, $y_1 = y_2$. Therefore $\mathbb{R}^2/H = \{(0, y) + H \mid y \in \mathbb{R}\}$. Now you should be able to find an isomorphism between this group and $\mathbb{R}$.