Is this an isomoprphism of groups?

**hmmmm** · July 29th 2011, 11:41 AM

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

If I have the following surjective homomorphism

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

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

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

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

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

Is this correct?

Thanks for any help

**girdav** · July 29th 2011, 01:08 PM

It's correct. M

**girdav** · July 29th 2011, 01:09 PM

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

**hmmmm** · July 29th 2011, 01:14 PM

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

**girdav** · July 29th 2011, 02:08 PM

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

**topspin1617** · August 2nd 2011, 06:11 AM

Uh,

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

**hmmmm** · August 2nd 2011, 06:47 AM

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

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

**hmmmm** · August 2nd 2011, 11:38 AM

What about:

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

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:

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

thanks for any help

**Drexel28** · August 2nd 2011, 12:55 PM

Originally Posted by hmmmm

What about:

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

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:

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

thanks for any help

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

**hmmmm** · August 2nd 2011, 01:41 PM

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

Thread: Is this an isomoprphism of groups?

LinkBack

Thread Tools

Display

Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Re: Is this an isomoprphism of groups?

Similar Math Help Forum Discussions

About minimal normal groups and subnormal groups

Abelian groups and Homology groups

Quotient Groups - Infinite Groups, finite orders

free groups, finitely generated groups

Order of groups involving conjugates and abelian groups

Search Tags