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$PGL(n, F)$ 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 ?

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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. – Gerry Myerson Mar 19 '12 at 9:01
So for example ${\rm PSL}(3,3) = {\rm PGL}(3,3)$. In any case, $|F|$ even makes no sense if $F$ is infinite. – Derek Holt Mar 19 '12 at 9:33
you are right I take mistake.I edit it – Babak Miraftab Mar 19 '12 at 13:39
I have read this assertion from – Babak Miraftab Mar 19 '12 at 13:48
up vote 3 down vote accepted

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.

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«[...] if and only if $H$ has the $n$th root in it» does not make sense. The $n$th root of what? – Mariano Suárez-Alvarez Mar 20 '12 at 0:42

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