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.

Solution 1:

The splitting field for the nontrivial irreducible representations of the cyclic group of order 5 over the field $F_4$ of order 4 is $F_{16}$, so all such representations have degree 2 over $F_4$.

So (using Maschke's Theorem), a nontrivial representation of a group of order 5 of degree 3 over $F_4$ must be the direct sum of a nontrivial irreducible of dimension 2 and the trivial module.

The centralizer of the image of this representation in ${\rm GL}_3(4)$ must fix both of these irreducible constituents. The centralizer of the constituent in ${\rm GL}_2(4)$ is cyclic of order 15, and the centralizer of the trivial constituent is just the scalar matrices - i.e. cyclic of order 3. So the full centralizer in ${\rm GL}_3(4)$ has order 45.

Since not all elements in this centralizer have determinant 1, its intersection with ${\rm SL}_3(4)$ has order 15. Since it contains the scalar matrices in ${\rm SL}_3(4)$, it follows that all elements of order 15 in ${\rm SL}_3(4)$ have fifth power equal to a scalar matrix, and so ${\rm PSL}_3(4)$ has no element of order 15.

I have just noticed that I have assumed that the inverse image in ${\rm SL}_3(4)$ of an element of order 15 in ${\rm PSL}_3(4)$ centralizes an element of order 5, but I will leave you to show that.

Solution 2:

Here is a proof without use of Maschke's theorem.

D.J.S. Robinson, A Course in the Theory of Groups, 2d edition, exerc. 3.2.6, p. 79 (before stating Maschke's theorem, which only happens in chapter 8), asks for a proof that $PSL(3, 4)$ has no element of order 15. The author gives the following hint : "Suppose there is such an element : consider the possible rational canonical forms of a preimage in $SL(3, 4)$."

I found a proof, but it doesn't use a rational canonical form and I don't see how a rational canonical form can help.

Here is my proof.

The center of $SL(3, 4)$ is formed by the scalar matrices in $SL(3, 4)$. (In this particular case, the scalar matrices in $SL(3, 4)$ are exactly the scalar matrices in $GL(3, 4)$, but this fact plays no role.) Thus the statement amounts to say that if $M$ is a matrix in $SL(3, 4)$ such that $M^{15}$ is scalar, then either $M^{3}$ or $M^{5}$ is scalar. Let $F$ denote the field with 4 elements. In this field, $-1 = 1$ and $2 = 0$. Let $a$ and $b$ denote the two elements of $F$ other than $0$ and $1$. Then $a^{2} + a + 1 = 0$, $b^{2} + b + 1 = 0$, $a^{3} = 1$, $b^{3} = 1$, $a + b = 1$, $a = b^{2}$, $b = a^{2}$, $ab = 1$. Let $X$ be a variable. We have

(1) $X^{3} - 1 = (X - 1) (X - a) (X - b)$.

Replacing $X$ by $X^{5} $, we get

(2) $X^{15} - 1 = (X^{5} - 1) (X^{5} - a) (X^{5} - b)$.

On the other hand,

(3) $X^{5} - 1 = (X - 1) (X^{2} + aX + 1) (X^{2} + bX + 1)$.

Replacing $X$ by $X/a^{2}$ in (3) gives

(4) $X^{5} - a = (X - b) (X^{2} + X + a) (X^{2} + aX + a)$.

Similarly,

(5) $X^{5} - b = (X - a) (X^{2} + X + b) (X^{2} + bX + b)$.

Bringing (3), (4) and (5) in (2), we get

(6) $X^{15} - 1 = (X - 1) (X^{2} + aX + 1) (X^{2} + bX + 1) (X - b) (X^{2} + X + a) (X^{2} + aX + a) (X - a) (X^{2} + X + b) (X^{2} + bX + b)$.

None of the four elements of $F$ is a root of the first polynomial of degree $2$ in the right member, thus this polynomial is irreducible over $F$. Since the hypotheses are symmetric in $a$ and $b$, the second polynomial of degree $2$ in the right member is also irreducible over $F$. In view of the manner in which the other polynomials of degree $2$ were obtained from the first two ones,

(7) all factors in the right member of (6) are irreducible polynomials over $F$.

Let $M$ be a matrix in $SL(3, 4)$ such that $M^{15}$ is scalar. We have to prove that either $M^{3}$ or $M^{5}$ is scalar.

Asume first that $M^{15} = 1$. Then $M$ annihilates the polynomial $X^{15} - 1$, thus, in view of (7), every irreducible factor of its characteristic polynomial is equal to one of the nine factors in the right member of (6). (Classical property of the characteristic polynomial.) Thus, since the characteristic polynomial of $M$ is of degree 3, it is equal either to $ (X - 1) (X - a) (X - b) $ or to the product of one of the three polynomials $X - 1$, $X - a$, $X - b$ by one of the six polynomials of degree 2 appearing in the scond member of (6). In the first case, the characteristic polynomial is equal to $X^{3} - 1$ (see (1) ), so $M^{3}$ is equal to 1 and is thus scalar. In the second case, note that the hypothesis $M \in SL(3, 4)$ implies $det(M) = 1$, thus the independent coefficient of the characterisic polynomial is $(-1)^{3} = 1$ in $F$. We deduce that the six possibilities for the characterisic polynomial are $(X-1) (X^{2} + aX + 1) $, $(X-1) (X^{2} + bX + 1) $, $(X-a) (X^{2} + X + b) $, $(X-a) (X^{2} + bX + b) $, $(X-b) (X^{2} + X + a) $, $(X-b) (X^{2} + aX + a) $. Thus, from (3), (4) and (5), the characterisic polynomial divides one of the three polynomials $X^{5} - 1$, $X^{5} - a$, $X^{5} - b$, thus $M^{5}$ is scalar.

Our thesis is thus right if $M^{15} = 1$. There remains to prove it in the case where, for example, $M^{15} = a$. This can be written $(M^{3}/a^{2})^{5} - 1 = 0$, thus, from (3),

(8) $M$ annihilates the polynomial $(X^{3} - a^{2} ) (X^{6} + X^{3} + a) (X^{6} + aX^{3} + a)$.

The only nonzero cube in $F$ is 1, thus the polynomial $X^{3} - a^{2}$ has no root in $F$. Since it is of degree $3$, it is irreducible over $F$. Let us prove that the polynomial $X^{6} + X^{3} + a$ is irreducible over $F$. Let $\theta$ denote a root of $X^{6} + X^{3} + a$. It suffices to prove that $F(\theta)$ is an extension of degree $6$ of $F$. Now, $F(\theta)$ is obtained by adjoining a cubic root of $\lambda$ to the field $F(\lambda)$, where $\lambda$ is a root of $X^{2} + X + a$. The norm of $\lambda$ in $F$ is $a$ and is thus not a cube in $F$, thus $\lambda$ is not a cube in $F(\lambda)$. Thus, since $X^{3} - \lambda$ is a polynomial of degree 3, it is irreducible over $F(\lambda)$, so $F(\theta)$ has degree 3 on $F(\lambda)$. On the other hand, we saw that the polynomial $X^{2} + X + a$ is irreducible over $F$, thus $F(\lambda)$ has degree $2$ over $F$. So, as announced, $F(\theta)$ is an extension of degree $6$ of $F$. As we saw, it proves that the polynomial $X^{6} + X^{3} + a$ is irreducible over $F$. Similarly, the polynomial $X^{6} + aX^{3} + a$ is irreducible over $F$. Thus (8) implies that every irreducible factor of the characteristic polynomial of $M$ is equal to one of the polynomials $X^{3} - a^{2}$, $X^{6} + X^{3} + a$, $X^{6} + aX^{3} + a$. But the characteristic polynomial of $M$ has degree $3$, thus it must be $X^{3} - a^{2}$, so $M^{3} = a^{2}$ and $M^{3}$ is scalar, which completes the proof.

Can the proof be simplified by use of a rational canonical form ? Thanks in advance for the answers.