1. ## Identifying Quotient Groups

Hello,

Given the set G of invertible, upper triangular 2x2 matrices, it can be verified that G forms a subgroup of the general linear group (invertible matrices) of 2x2 matrices with real entries.

Now, consider the set H of matrices of the form

$\displaystyle H = \left(\begin{array}{cc}1&b\\0&c\end{array}\right)$

where $\displaystyle b,\ c$ are real and $\displaystyle c \neq 0$

Now, it can be verified that H is a subgroup of G
In fact, it can be verified (I have done the dirty work!) that H is normal in G.

So, now we are asked to identify the quotient group $\displaystyle G/H$

That is to say, we are asked to use the First Isomorphism Theorem to show it is isomorphic to a known group.

So by my understanding, the trick is to identify some surjective homomorphism $\displaystyle \Phi$ whose kernel is the group H, $\displaystyle ker(\Phi) = H$, and then the image of that homomorphism, $\displaystyle Im(\Phi)$, is isomorphic to the quotient group $\displaystyle Im(\Phi) \approx G/H$?

Any help appreciated, thanks!!

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!

3. Originally Posted by matt.qmar

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!
Looks good to me, as long as you do prove that that is a homomorphism.

You have a normal subgroup, N, and it is the kernel of a homomorphism. Thus, apply the first isomorphism theorem to get that,

$\displaystyle G/N=G/Ker(\phi) \cong im(\phi) = \mathbb{R}$.