10

I want to classify the center of the Special Linear Group. I already determined the center for $SL(n,F)$:

$$Z(SL(n,F))=\left\{ \lambda I_n:\lambda^n=1 \right\}.$$

I showed that $Z(SL(n,F))$ is itself a group and now

I want to show that $Z(SL(n,F))$ is cyclic and has a order dividing $n$.

I thought this is possible by regarding the map $SL(n,F)\rightarrow P(F)$, $P(F)$ being the projective space. The kernel of this map is $Z(SL(n,F))$. How do I have to argue now?

José Carlos Santos
  • 427,504
johnka
  • 569

1 Answers1

8

You have already proved that $Z\bigl(SL(n,F)\bigr)$ is (isomorphic to) the group $\{\lambda\in F\mid\lambda^n=1\}$. And this group is cyclic, since finite subgroups of the multiplicative group of a field are cyclic. Extending $F$ to its algebraic closure $\overline F$, you'll get a group with $n$ elements (the $n$ roots of the polynomial $x^n-1$). Since your group is a subgroup of this one and this one has order $n$, the order of your group divides $n$, by Lagrange's theorem.

user26857
  • 52,094
José Carlos Santos
  • 427,504
  • I dont quit understand how we get the isomorphism? – johnka Jan 01 '18 at 21:15
  • @johnka The isomorphism is provide by the map$$\begin{array}{ccc}{\lambda\in F,|,\lambda^n=1}&\longrightarrow&Z\bigl(SL(n,F)\bigr)\\lambda&\mapsto&\lambda\operatorname{Id}_n.\end{array}$$ – José Carlos Santos Jan 01 '18 at 21:34
  • 3
    If $char\ F=p>0$, we can't say the $n$ roots of the polynomial $x^n-1$ in $\overline F$ has $n$ (distinct) elements. Rather, write $n=p^rm, (p, m)=1$, then there are exactly $m$ elements. – Lao-tzu Nov 10 '18 at 17:03