Originally Posted by

**matt.qmar** Some quick progrss....

If we define a homomorphism $\displaystyle \Phi: G \rightarrow R$ for the reals under multiplication $\displaystyle R$, by

$\displaystyle \Phi(\left(\begin{array}{cc}a&b\\0&c\end{array}\ri ght)) = a$

(I have not checked that this is actually operation preserving but I think it is...)

Then the identity of $\displaystyle R$ is 1, so the $\displaystyle ker(\Phi) = \left(\begin{array}{cc}1&b\\0&c\end{array}\right)$

which is just the group H.

Since $\displaystyle \Phi$ maps $\displaystyle G$ onto $\displaystyle R$

Then by the First Isomorphism Theorem, do we have $\displaystyle G/H \approx R$??

Not sure if this is solid... A second opinion would be helpful. Thanks!