5
$\begingroup$

$PGL(n, F)$ and $PSL(n, F)$ are equal if and only if every element of $F$ has an $nth$ root in $F$.($F$ is finite field)

I can show that if $PGL(n, F)=PSL(n, F)$ then $|F|$ have to be even.I have not any idea how to deal with it. any suggestions ?

  • 2
    If $F$ is the field of 3 elements, then every element of $F$ has a cube root in $F$, so I don't understand how you get that the order of $F$ is even.2012-03-19
  • 0
    So for example ${\rm PSL}(3,3) = {\rm PGL}(3,3)$. In any case, $|F|$ even makes no sense if $F$ is infinite.2012-03-19
  • 0
    you are right I take mistake.I edit it2012-03-19
  • 0
    I have read this assertion from http://en.wikipedia.org/wiki/Projective_linear_group2012-03-19

1 Answers 1

4

The determinant map gives rise to a split exact sequence

$$ PSL(n, F) \hookrightarrow PGL(n,F) \twoheadrightarrow F^\times / (F^\times)^n, $$

i.e. we have an isomorphism $PSL(n,F) \rtimes F^\times / (F^\times)^n \cong PGL(n,F)$.

Now, $F^\times / (F^\times)^n = 1$, if and only if every element of $F$ has a $n$-th root.