# Thread: center of general linear group

1. ## center of general linear group

I am trying to show that Z(GL(n,F*)) $\displaystyle \cap$ SL(n,F*) $\displaystyle \cong$ T(n) ... This is for any non-zero Field

Where Z is the center of General Linear Group.
and T(n)={a^n=1 | a $\displaystyle \in$ F*}

Im trying to figure this out first Z(GL(n,F*)) SL(n,F*), since the determinant in this set is suppose to be 1, then det(aI) = det(a)det(I) = det(a), this has to be 1 to be in the set therefore a=1, which implies this set contains I(n,F*) (The identity matrix) only.

Am I right so far???

Don't really know what else to do.

2. Originally Posted by Dreamer78692
I am trying to show that Z(GL(n,F*)) $\displaystyle \cap$ SL(n,F*) $\displaystyle \cong$ T(n) ... This is for any non-zero Field

Where Z is the center of General Linear Group.
and T(n)={a^n=1 | a $\displaystyle \in$ F*}

Im trying to figure this out first Z(GL(n,F*)) SL(n,F*), since the determinant in this set is suppose to be 1, then det(aI) = det(a)det(I) = det(a), this has to be 1 to be in the set therefore a=1, which implies this set contains I(n,F*) (The identity matrix) only.

Am I right so far???

Don't really know what else to do.

You really need to show that $\displaystyle Z:=Z\left(GL(n,\mathbb{F})\right)=\{aI\,|\,a\in\ma thbb{F}^*\}$, and then

all falls in place.

You can show the above by taking $\displaystyle B\in Z$ and checking what the coefficients of B must be by

analyzing $\displaystyle BE_{ij}=E_{ij}B\,,\,\,with\,\,E_{ij}=$ the matrix with 1 in the ij entry and zero everywhere else.

Tonio

Pd Pay attention to the notation of the general linear group: it is not F* in it but F...

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# center of a general linear group

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