Given the set G=\begin{bmatrix}a&b\\0&c\end{bmatrix}, where a,c ∈R*, b ∈R and its subgroup H=\begin{bmatrix}a&b\\0&a^m\end{bmatrix}, where a ∈R*, b ∈R and m ∈Z, show that the factor group G/H is isomorphic to the multiplicative group of real numbers.
I started with proving that H is normal subgroup of G, therefore its left and right cosets are equal. I tried using Theorem on homomorphisms which says that if ϕ(G) -> R is homomorphism and ker(ϕ)=H, then the factor group G/H is isomorphic to (R,*).
My idea was to define homomorphism between determinants and real numbers, such that if ϕ(det(g1)*(det(g2)) = r1*r2 = ϕ(det(g1))*ϕ(det(g2)), where g1,g2 ∈G and r1,r2 ∈R. The next step is to show that ker(ϕ)=H. But I don't see a way to show that a number such as a*a^m is mapped into the neutral element(which is 1 in real numbers). So maybe, my approach is not correct but I don't have any clue of another.
Looking forward for your advice!
Best Answer
Try with the function \begin{align*} f \ : \ G \quad & \longrightarrow \mathbb{R}^\times \\ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} & \longmapsto \frac{a^m}{c} . \end{align*}