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$PGL(n, F)$ and $PSL(n, F)$ are isomorphic if and only if every element of $F$ has an $n$th root in $F$. ($F$ is a 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?

user26857
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Bobby
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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 09:01
  • 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 09:33
  • I have read this assertion from http://en.wikipedia.org/wiki/Projective_linear_group – Bobby Mar 19 '12 at 13:48

1 Answers1

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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.

Marc Palm
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    Sorry, I still have difficulty understanding the other direction, if $F^\times /(F^\times)^n \neq 1$, how can we know that $PSL(n,F)$ is not isomorphic to $PGL(n,F)$? – Tsoshamry Feb 19 '24 at 03:29
  • If F is a finite field, then the cardinalities don't match and hence they can't be isomorphic unless $F^{}/(F^{})^{n} = 1$ – Angry_Math_Person May 01 '24 at 19:20