Equality between two definitions of the projective special unitary group

Question

Let $q$ be a prime power and let $F=GF(q^2)$ be the Galois field of $q$ elements. Then $$((x_1,x_2,x_3),(y_1,y_2,y_3))\longmapsto x_1y_3^q + x_2y_2^q + x_3y_1^q$$ is a Hermitian form. The subset of the general linear group $GL(3,q^2$) preserving this bilinear form is called the general unitary group $GU(3,q)$. The subset of $GU(3,q)$ of determinant 1 is called the special unitary group $SU(3,q)$.

I encountered two definitions for the projective special unitary group $PSU(3,q)$.

1. $PSU(3,q) = \frac{SU(3,q)Z}{Z} = \frac{(GU(3,q) \cap SL(3,q^2))Z}{Z} $.
2. $PSU(3,q) = PGU(3,q)\cap PSL(3,q^2) = \frac{GU(3,q)Z}{Z} \cap \frac{SL(3,q^2)Z}{Z} $.

Here, $Z=\left\{ kI:0\neq k\in GF(q^2) \right\}$ is the center of $GL(3,q^2))$.

The first definition says that the members of $PSU(3,q)$ are cosets $AZ$ of $Z$ such that $A$ is a multiple of a matrix that is both unitary and special linear. This definition obtained from the second isomorphism theorem and $SU(3,q)/(SU(3,q)\cap Z)=SU(3,q)/(\operatorname{center}(SU(3,q)))$.

The second definition says that the members of $PSU(3,q)$ are cosets $BZ$ of $Z$ such that $B$ is a multiple of a matrix that is unitary and a multiple of a matrix that is special linear. This definition is from the book Permutation Groups by Dixon and Mortimer.

Clearly, containment in the first implies containment in the second. However, I cannot figure out how to show the containment in the other direction: namely, if $B=kU=jS$ for some unitary $U$ and special linear $S$, then there is a scalar $l$ and a special unitary matrix $T$ such that $B=lT.$

Answer 1

Here is an argument to prove what you want, but I have the feeling that there is a more conceptual and illuminating proof. I will do the argument for dimension $3$ but it should be in all dimensions.

The problem seems to be the following. Let $g \in {\rm GU}(3,q)$ and suppose that there is a scalar matrix $\lambda I_3$ with $\det(\lambda g) = 1$. We need to prove that there exists such a $\lambda$ with $\lambda I_3 \in {\rm GU}(3,q)$, that is, with $\lambda^{q+1} = 1$.

Let $d = \det(g)$. Then $g$ unitary implies $d^{q+1} = 1$. Or, using logs with respect to a primitive element $\omega$ of ${\rm GF}(q^2)$, we have $\log(d) = (q-1)k$ for some $k$.

Now $1 = \det (\lambda g) = \lambda^ 3 d$, so $d$ is a cube and hence $3|(q-1)k$, and the three cube roots of $d$ are $\frac{(q-1)k}{3} + t\frac{q^2-1}{3}$ for $t=0,1,2$.

Now, if $3$ does not divide $q-1$, then it divides $k$, and $\lambda = \omega^{-\frac{(q-1)k}{3}}$ satisfies $\lambda^{q+1}=1$.

Otherwise $3|(q-1)$ and the cube roots of $d$ are $\frac{q-1}{3}\left(k + t\frac{q+1}{3}\right)$ for $t=0,1,2$, and in this case, since $3$ does not divide $q+1$, there is a value of $t$ such that $k + t\frac{q+1}{3}$ is divisible by $3$, and again we can find $\lambda$ with $\lambda^3=1$.