Show that Z(SL(n,F)) $\displaystyle \cong$ {a $\displaystyle \in$ F* | a^n = 1}.
I know that if X $\displaystyle \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.
Goku: What you've written doesn't quite make sense. The center will $\displaystyle \{aI_n : a^n=1\}$.
Clearly this set is contained in the center. Conversely, let $\displaystyle A=(a_{i j})$ be in the center. Let $\displaystyle E_{i j}$ be the matrix with 1 in the $\displaystyle (i,j)$th position and 0 elsewhere. Then $\displaystyle I_n+E_{i j}\in SL(n, F)$ if $\displaystyle i\ne j$. Now use the fact that this matrix must commute with $\displaystyle A$.