<s>[INST] Show that $PSL(3,4)$ has no element of order $15$. 
$PSL(3,4)$ has no element of order $15$. Thus it is no isomorphic to $A_8$.

Here, $PSL(3,4)$ denotes the $3 \times 3$ projective special linear group on the field with $4$ elements.
As listing all the elements takes too much work, is there any better way to prove there is no element of order $15$ in $PSL(3,4)$? 
Thank you very much.
 [/INST] I try to write out the proof sketched by user641.
Let $F$ denote the field with 4 elements. Speaking of a polynomial, "irreducible" will mean "irreducible over $F$".
We have to prove that $PSL(3, 4)$ has no element of order 6 and this amounts to prove that if $M$  is a matrix in $SL(3, 4)$ such that $M^{6}$ is scalar, then either $M^{2}$ or $M^{3}$ is scalar.
Assume first that $M^{6} = 1$. Then $M$ annihilates the polynomial $X^{6} - 1 = (X-1)^{2} (X-a)^{2} (X-b)^{2}$, where $a$ and $b$ denote the two nonzero elemets of $F$ other than $1$. Thus the characteristic polynomial of $M$ is of the form $(X-c) (X-d) (X-e)$, with $c$, $d$ and $e$ nonzero in $F$. Since the determinant of $M$ is $1$, the characteristic polynomial of $M$ must be equal to $(X-1) (X-a) (X-b) = X^{3} - 1$ or of the form $(X-f)^{3}$ with $f$ nonzero in $F$. In the first case $M^{3} = 1$, thus $M^{3}$ is scalar. In the second case, $M$ annihilates the polynomial $(X-f)^{3} = X^{3} + fX^{2} + f^{2}X + f^{3}$, thus $M^{4} = f^{4}$. Since $M^{6} = 1$, we have $M^{2} =f^{-4} = f^{2}$, thus $M^{2}$ is scalar.
Thus our thesis is true in the case where $M^{6} = 1$. There remains to prove it when $M^{6} = a$, where $a$ is one of the two nonzero elements of $F$ different from $1$. We have $a^{3} = 1$, thus $a^{4} = a$, thus the polynomial $X^{6} - a$ is equal to $X^{6} - a^{4} = (X^{3} - a^{2})^{2}$. So $M$ annihilates the polynomial $(X^{3} - a^{2})^{2}$. Since $X^{3} - a^{2}$ is irreducible, this implies that the only irreducible factor of the characteristic polynomial of $M$ is equal to  $X^{3} - a^{2}$. Since the characteristic polynomial of $M$ is of degree 3, it must be equal to $X^{3} - a^{2}$, so $M^{3} = a^{2}$, so $M^{3}$ is scalar, which completes the proof.
Again, I don't see the usefulness of the rational canonical form.
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