# Is $G/N$ isomorphic to $\mathbb R ?$

$$G=\left\{\begin{bmatrix} a & b \\ 0 & \ \ \ \ a^{-1}\end{bmatrix} : a,b \in \mathbb R;a>0 \right\}$$

$$\hspace{-1.1in}N=\left\{\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} :b\in \mathbb R\right\}$$

Which of the following are true?

(A) $G/N$ is isomorphic to $\mathbb R$ under addition.

(B) $G/N$ is isomorphic to $\{a ∈ \mathbb R : a > 0\}$ under multiplication.

(C) There is a proper normal subgroup $N'$ of $G$ which properly contains $N$.

(D) $N$ is isomorphic to $\mathbb R$ under addition.

Option $D$ is not true I figured. Option $C$ is possible as every subgroup is contained in a normal subgroup that is the normalizer . But how can I tel if that one is a proper subgroup or not ?

Also more than one correct answers is a possibility and I'm not sure about options $A$ and $B.$

Please give me hints. I think I can work out from thereon.

Thanks.

• @FaraadArmwood : thank you for the edit. – user118494 Apr 27 '16 at 8:17
• No problem. Try to add some work that you've done. – Faraad Armwood Apr 27 '16 at 8:20
• What made you decide $N$ is not isomorphic to {the reals under addition}? Did you try multiplying two elements of $N$ to see what you get? – Gerry Myerson Apr 27 '16 at 8:55
• Are you still here? – Gerry Myerson Apr 28 '16 at 12:41

Try to show that the map $$\begin{pmatrix} a & b \\ 0 & a^{-1} \\ \end{pmatrix} \mapsto a$$ is a surjective homomorphism from $G$ to $(\mathbb R^+,\cdot)$.
If it helps, you can have a look at a similar answer here: Proving that $G/N$ is an abelian group