0
$\begingroup$

So i have the set $G=\left\{ \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}\mid a,b,c \in \mathbb{R}\right\}$.

I proved already that it's a group together with multiplication of matrices.

So now i have to find this groups centre:

$Z(G) =\{x \in G \mid xg=gx, \forall g \in G \}$.

So after that i need to specify to what known group that centre is isomorphic to and prove that statement. So i know i will have find an isomorphism between $Z(G)$ and the mystery group, but i think the hardest part for me is finding which elements are in centre.

Any help would be appreciated.

  • 4
    Why dont you brute force it? Take two such matrices, multiply them both ways and compare the results. – 2017-02-07
  • 2
    You can always take a generic matrix $g(a,b,c)$ and find conditions on a,b,c so that $g$ commutes with a generic matrix $h=h(x,y,z)$, and then find those a,b,c that satisfy the conditions for all x,y,z. Another approach (which I think is much better) is to find generators for your group, and check which matrices commutes with all of the generators. – 2017-02-07
  • 1
    @Ofir: your first approach works fin and neccitates only two matrix mutriplications. The result is remarkable. – 2017-02-07

1 Answers 1

0

Note that any matrix in your group can be written as $$ \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & b-ac \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ Letting $E_{i,j}$ be the zero matrices with 1 in the (i,j) coordinate, you see that you group is generated by elements of the form $I+rE_{i,j}$ where $r$ is real and $1\leq i

For the first type of generator you need to check that $[g,I+rE_{i,j}]=e$ or equivalently $$(I+xE_{1,2}+yE_{1,3}+zE_{2,3})(I+rE_{i,j})=(I+rE_{i,j})(I+xE_{1,2}+yE_{1,3}+zE_{2,3})$$ and simplifying $$(xE_{1,2}+yE_{1,3}+zE_{2,3})rE_{i,j}=rE_{i,j}(xE_{1,2}+yE_{1,3}+zE_{2,3})$$ (You should do this in matrix notation also - I use this notation because it is easier to write it like this and it can be easily generalized to matrices of larger dimension).

Taking for example $(i,j)=(1,2)$ you get that $0=rzE_{1,3}$ and since this should be true for all $r$, you must have that $z=0$. Similarly, taking $(i,j)=(2,3)$ you get that $x=0$. Finally, taking $(i,j)=(1,3)$ (and assuming that $x=z=0$) you need to check that both sides of the equation are 0, so that your center is exactly the matrices of the form $I+yE_{1,3}$.