Throughout, let $q$ be an odd prime power. Let $GF(q^2)$ be the field with $q^2$ elements.
My question concerns the generation of the special unitary group $SU(n,q^2)$, where $n \ge 2$, by certain subgroups isomorphic to $SU(2,q^2)$. Before asking the question, I give some background definitions which should help to understand and answer the question. Those who are familiar with the definition of special unitary groups over finite fields can skip the background definitions.
Background Definitions and Remarks
Definition: Let $V$ be a finite-dimensional vector space over $GF(q^2)$. A hermitean form on $V$ is a map $f: V \times V \rightarrow GF(q^2)$ satisfying the following:
- $f(u+v,w) = f(u,w)+f(v,w)$ for all $u,v,w \in V$,
- $f(\lambda v, w) = \lambda f(v,w)$ for all $v,w \in V$, $\lambda \in GF(q^2)$,
- $f(v,w) = f(w,v)^q$ for all $v,w \in V$.
A hermitean form $f$ on a finite-dimensional vector space over $GF(q^2)$ is called nondegenerate if whenever $v$ is an element of $V$ with $f(v,w) = 0$ for all $w \in V$, we have $v = 0$. A unitary space over $GF(q^2)$ is a finite-dimensional vector space $V$ over $GF(q^2)$ together with a nondegenerate hermitean form on $V$.
Remark: Let $n \ge 1$. Then it can be shown that there is, up to isomorphism, precisely one $n$-dimensional unitary space over $GF(q^2)$. The existence is quite easy to see. For the uniqueness, one needs the fact that each unitary space over $GF(q^2)$ possesses an orthonormal basis ([2, Theorem 10.3]) and the observation that any two $n$-dimensional unitary spaces over $GF(q^2)$ are isomorphic whenever both of them have an orthonormal basis.
Definition: Let $n \ge 1$ and let $V$ be an $n$-dimensional unitary space over $GF(q^2)$. Let $f$ be the associated hermitean form on $V$. An isometry of $V$ is a bijective linear transformation of $V$ which preserves $f$. The group of all isometries of $V$ with determinant $1$ is said to be the special unitary group $SU(n,q^2)$.
By the above remark, $SU(n,q^2)$ is, up to isomorphism, independent of the choice of $V$.
Remark: With respect to an orthonormal basis of $V$, the group $SU(n,q^2)$ consists of all matrices $A$ in $SL(n,q^2)$ satisfying $\overline A A^t = I_n$, where $\overline A$ denotes the matrix obtained from $A$ by replacing each entry $a_{ij}$ by $a_{ij}^q$ and where $A^t$ is the transpose of $A$ ([2, p. 93]).
The question
Let $n > 2$. Recall that $q$ denotes an odd prime power. Let us consider $SU(n,q^2)$ as a group of matrices as described in the preceeding remark.
In the introduction of [1], the following fact is mentioned:
Let $G := SU(n+1,q^2)$ and let $U_i \cong SU(2,q^2)$, where $1 \le i \le n$, be the subgroups of $G$ corresponding to the $2 \times 2$ blocks along the main diagonal, i.e. $U_i$ is the subgroup of $G$ consisting of the matrices $A \in G$ such that if $1 \le k,l \le n+1$, then the $(k,l)$-entry of $A$ can only be nonzero if $k = l$, $(k,l) = (i,i+1)$ or $(k,l) = (i+1,i)$. Then $G$ is generated by the subgroups $U_1, \dots, U_n$.
My question is how the validity of this statement can be seen. I would also be happy with a reference containing a proof of this statement. If one replaces $SU(n+1,q^2)$ by $SL(n+1,q)$, then the resulting statement can be proved by induction using the fact that $SL(n+1,q)$ is generated by elementary matrices. But I do not see how a similar approach could work for $SU(n+1,q^2)$.
References:
[1] Bennett, Curtis D. and Shpectorov, Sergey, A new proof of a theorem of Phan, J. Group Theory 7 (2004), no. 3, 287–310, available at https://digitalcommons.lmu.edu/cgi/viewcontent.cgi?article=1000&context=math_fac.
[2] Grove, Larry C.: Classical groups and geometric algebra, Graduate Studies in Mathematics, 39. American Mathematical Society, Providence, RI, 2002.