$PGL(n, F)=PSL(n, F)$

matrices abstract-algebra group-theory

$PGL(n, F)$ 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 ?


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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