Thread:

1. center of SL(n,F)

Show that Z(SL(n,F)) $\cong$ {a $\in$ F* | a^n = 1}.

I know that if X $\in$ Z(SL(n,F)) then det(X) = det(aI) = a^n which has to equal 1 since it is in SL(n,F).

Any Help greatly appreciated.

2. Hi!

You write X=aI, so i suppose you know that a matrix commutes with any other matrix iff it is a multiple of the identity. The same holds for SL(n,F).

Banach

3. if X Z(SL(n,F)) then det(X) = det(aI) = a^n which has to equal 1 since it is in SL(n,F).
So does this count as a proof.

4. Goku: What you've written doesn't quite make sense. The center will $\{aI_n : a^n=1\}$.

Clearly this set is contained in the center. Conversely, let $A=(a_{i j})$ be in the center. Let $E_{i j}$ be the matrix with 1 in the $(i,j)$th position and 0 elsewhere. Then $I_n+E_{i j}\in SL(n, F)$ if $i\ne j$. Now use the fact that this matrix must commute with $A$.
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