# How the quotient group of following matrix group looks like?

Consider the following subsets of the group of $2\times 2$ non-singular matrices over $\mathbb{R}$: $G=\left\{ \begin{bmatrix}a&b\\ 0&d\end{bmatrix}: a,b,d\in \mathbb{R}, ad=1\right\}$

$H=\left\{ \begin{bmatrix}1&b\\ 0&1\end{bmatrix}: b\in \mathbb{R}\right\}$

Which of the following statements are correct?

1. $G$ forms a group under matrix multiplication.
2. $H$ is a normal subgroup of $G$.
3. The quotient group $G/H$ is well defined and is abelian.
4. The quotient group $G/H$ is well defined and is isomorphic to the group of 2x2 diagonal matrices (over reals) with determinant 1.

So I checked that option 1 is true. Also for all $g\in G$, $g^{-1}Hg=H$ so $H$ is a normal subgroup of $G$. Option 2 is also correct. But I am stuck with option 3 and 4. Since $H$ is normal, we can talk about the quotient group $G/H$, but I don't know how this quotient group is look like ? and also the well define part. Also I need help with the 4th option. How can I do this? Any help would be great. thanks.

Hint:

Let $K$ be the group of $2\times 2$ diagonal matrices (over reals) with determinant $1$.

Define $f:G\rightarrow K$ by $$\begin{pmatrix}a&b\\0&d\end{pmatrix}\mapsto \begin{pmatrix}a&0\\0&d\end{pmatrix}$$ Try to verify that this is a surjective homomorphism with kernel $H$.
Then apply the first isomorphism theorem.

A coset of $H$ looks like

$$\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} H= \left\{ \begin{bmatrix} a & b \\ 0 & d\end{bmatrix}\begin{bmatrix} 1 & h \\ 0 & 1 \end{bmatrix} : ad=1 \right\}$$

Multiply out

$$\begin{bmatrix} a & b \\ 0 & d\end{bmatrix}\begin{bmatrix} 1 & h \\ 0 & 1 \end{bmatrix}=\begin{bmatrix} a & ah+b \\ 0 & d\end{bmatrix}$$

Can you think of an $h$ that makes this matrix particularly nice? If so, you could create a very nice set of coset representatives. In particular, a set of coset representatives which is itself a subgroup.

Proposition. If $G$ is a group and $H,K$ are subgroups such that $H$ is normal and $K$ is a set of coset representatives for $H$, then $G/H\cong K$.

• I didn't get what you mean by "particularly nice"? If I choose $h=-b/a$ then it turnout to be a diagonal matrix. Is that what you mean? – Kushal Bhuyan Mar 23 '17 at 10:08
• Yes. ${}{}{}{}$ – arctic tern Mar 23 '17 at 15:46
• OK. What about the "well define" thing? How to check that? – Kushal Bhuyan Mar 24 '17 at 1:57
• Also can you refer me a proof to the above-mentioned proposition? – Kushal Bhuyan Mar 24 '17 at 2:02
• I'd ignore the "well-defined" part. They're presumably trying to say that $g_1H\cdot g_2H=g_1g_2H$ actually makes sense so that $G/H$ is a group, which follows from $H$ being normal (if $H$ were not normal that description would not actually define a function $G/H\times G/H\to G/H$). As for the proposition I gave, you should prove it yourself. It's good practice and good for understanding what's going on. – arctic tern Mar 24 '17 at 3:24