Is this an isomoprphism of groups?

G is the group of 2x2 invertible upper triangular matrices with real entries. Let H be the normal subgroup such that $\displaystyle a_{11}=a_{22}=1$.

If I have the following surjective homomorphism

$\displaystyle f:G\rightarrow (\mathbb{R}^*,\times)$

such that $\displaystyle f(g)=a_{11}a_{22}$.

Then by the first isomorphism theorem I have the following isomorphism:

$\displaystyle f:G/H\rightarrow (\mathbb{R}^*,\times)$

such that $\displaystyle f(\bar{g})=a_{11}a_{22}$

Is this correct?

Thanks for any help

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

It's correct. Maybe you should denote by $\displaystyle \widetilde f$ the morphism between $\displaystyle G/H$ (I guess $\displaystyle H$ is the matrix of $\displaystyle G$ which have determinant $\displaystyle 1$) and $\displaystyle \mathbb R^*$.

Re: Is this an isomoprphism of groups?

yeah thanks very much, I edited it. My notation is bad, how did you put the squiggle above the f?

Re: Is this an isomoprphism of groups?

Quote:

Originally Posted by

**hmmmm** yeah thanks very much, I edited it. My notation is bad, how did you put the squiggle above the f?

\widetilde f

Re: Is this an isomoprphism of groups?

Uh,

I don't think so. The kernel of that homomorphism is certainly strictly larger than $\displaystyle H$. What about the matrix $\displaystyle A= \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$? Certainly $\displaystyle A\in G\setminus H$ but $\displaystyle \mathrm{det}A=1$.

Re: Is this an isomoprphism of groups?

Ah of course, in fact if $\displaystyle a_{22}=(a_{11})^{-1}$ then that maps to the identity.

So i have to have another map f such that $\displaystyle f(a_{11},a_{22})=1$ and it has to be surjective right?

Re: Is this an isomoprphism of groups?

What about:

Let J be the subgroup of $\displaystyle \mathcal{SL}_r(\mathbb{R})$ such that $\displaystyle a_{12}=a_{21}=0$.

Then we have a surjectrive homomorphism [\bar{f}:G\rightarrow J[/TEX] with kernel H.

So from the first isomorphism theorem we have an isomorphism:

$\displaystyle \widetilde{f}:G/H\rightarrow J$

thanks for any help

Re: Is this an isomoprphism of groups?

Quote:

Originally Posted by

**hmmmm** What about:

Let J be the subgroup of $\displaystyle \mathcal{SL}_r(\mathbb{R})$ such that $\displaystyle a_{12}=a_{21}=0$.

Then we have a surjectrive homomorphism [\bar{f}:G\rightarrow J[/TEX] with kernel H.

So from the first isomorphism theorem we have an isomorphism:

$\displaystyle \widetilde{f}:G/H\rightarrow J$

thanks for any help

What precisely are you trying to do? Find a nicer description of $\displaystyle G/H$? Is this a different question?

Re: Is this an isomoprphism of groups?

Yeah that is what I am trying to do, any help would be apperciated