# Thread: Centre of a group

1. ## Centre of a group

Could anyone help me please? I don't have any examples of how to work out the centre of a group, so I'm struggling. I keep trying to apply the definition but any way I try doesn't seem to work.

Work out the centre of

$\displaystyle T(2,R)= \{ \left(\begin{array}{cc}a&b\\0&c\end{array}\right); a,b,c$ elements of $\displaystyle R, ac \neq 0 \}$

The $\displaystyle R$ should be the symbol representing the real numbers but I don't know how to put that in.

Thanks.

2. Originally Posted by Nightfly
Could anyone help me please? I don't have any examples of how to work out the centre of a group, so I'm struggling. I keep trying to apply the definition but any way I try doesn't seem to work.

Work out the centre of

$\displaystyle T(2,R)= \{ \left(\begin{array}{cc}a&b\\0&c\end{array}\right); a,b,c$ elements of $\displaystyle R, ac \neq 0 \}$

The $\displaystyle R$ should be the symbol representing the real numbers but I don't know how to put that in.
You want to find, $\displaystyle A,B,C$ so that,
$\displaystyle \begin{bmatrix}A&B\\C&0 \end{bmatrix} \begin{bmatrix} a & b \\ c & 0 \end{bmatrix} = \begin{bmatrix} a & b \\ c & 0 \end{bmatrix} \begin{bmatrix}A&B\\C&0 \end{bmatrix} \text{ for all }a,b,c$.

3. Yeah, that's what I thought except I got

$\displaystyle \left(\begin{array}{cc}a&b\\0&c\end{array}\right)\ left(\begin{array}{cc}A&B\\0&C\end{array}\right)=\ left(\begin{array}{cc}aA&aB+bC\\0&cC\end{array}\ri ght)$

$\displaystyle \left(\begin{array}{cc}A&B\\0&C\end{array}\right)\ left(\begin{array}{cc}a&b\\0&c\end{array}\right)=\ left(\begin{array}{cc}aA&Ab+Bc\\0&cC\end{array}\ri ght)$

so

$\displaystyle aB+bC=Ab+Bc$

which gives

$\displaystyle (a-c)B-(A-C)b=0$

but then I couldn't figure out how to get the values for $\displaystyle A,B,C$

4. Originally Posted by Nightfly
$\displaystyle aB+bC=Ab+Bc$
This equation is satisfied for all $\displaystyle a,b,c\in \mathbb{R}, ac\not = 0$.
If this works for any values then it works for $\displaystyle b=0$ and we get,
$\displaystyle aB = Bc \implies (a-c)B = 0 \implies B=0$ since $\displaystyle a-c\not = 0$ for all $\displaystyle a,c$.
But if $\displaystyle B=0$ then it forces $\displaystyle A=C$.

Thus the center is the set, $\displaystyle \left\{ \begin{bmatrix} t&0\\0&t \end{bmatrix} : t\in \mathbb{R}^{\times} \right\}$.