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

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

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.


  • $\begingroup$ @FaraadArmwood : thank you for the edit. $\endgroup$ – user118494 Apr 27 '16 at 8:17
  • $\begingroup$ No problem. Try to add some work that you've done. $\endgroup$ – Faraad Armwood Apr 27 '16 at 8:20
  • $\begingroup$ 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? $\endgroup$ – Gerry Myerson Apr 27 '16 at 8:55
  • $\begingroup$ Are you still here? $\endgroup$ – 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 $(\mathbb R^+,\cdot)$.

What is kernel of this homomorphism? What can you say based on this using first isomorphism theorem?

This should help you to see that (B) is true.

If it helps, you can have a look at a similar answer here: Proving that $G/N$ is an abelian group


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