Matrix groups

1. Jan 31, 2008

Mattofix

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

I have done part a, proving the four axioms. I just dont understand part b) if someone could point me in the right direction...

2. Jan 31, 2008

Dick

Suppose g=[a b,0 d] (I hope the informal matrix notation is clear) is in the center. That means for ANY g'=[a' b',0 d'] in the group, that g*g'=g'*g. Write that out in components. You should find a*b'+b*d'=a'*b+b'*d. Now since the primed components are arbitrary, what does this imply for g if you pick say, a'=1, b'=0 and d'=0. Or if you pick a'=0, b'=1 and d'=0???

3. Feb 1, 2008

Mattofix

a = d , b = 0 ?

4. Feb 1, 2008

Mattofix

...so Z(G) = ( a 0 ,0 a ) ?

5. Feb 1, 2008

HallsofIvy

Staff Emeritus
Does that say "Z(G)= {g | gh= hg for all h}"? The lower bar of the "=" is very pale and it looks like "gh- hg" which makes no sense!

6. Feb 1, 2008

Mattofix

yes it does - strange though, shows up fine on my screen

7. Feb 1, 2008

Mattofix

a quick reply would be much appreciated as i have to hand this in pretty soon

8. Feb 1, 2008

Dick

Yes, the center is the multiples of the identity.

Last edited: Feb 1, 2008
9. Feb 1, 2008

thanks