Remark

If D is a field with two elements, then $\mathbf{G}\mathbf{L}_2(D)$ is equal to $\mathbf{S}\mathbf{L}_2$(D), and it is a group of order 6 whose derived group has order 3. If D is a field with three elements, then the group $\mathbf{S}\mathbf{L}_2(D)$ has order 24, and its derived group has order 8. Suppose $n\geqslant 2$; except in the two previous cases, every normal subgroup of $\mathbf{S}\mathbf{L}_n(D)$ distinct from $\mathbf{S}\mathbf{L}_n(D)$ is contained in the center $Z(\mathbf{S}\mathbf{L}_n(D))$ of $\mathbf{S}\mathbf{L}_n(D)$ (II, §10, p. 421, Exercises 13 and 14). The group $\mathbf{S}\mathbf{L}_n(D)/Z(\mathbf{S}\mathbf{L}_n(D))$ is then simple.

Let V be a right vector space over the field D, of finite dimension $n$. We denote by SL(V), and call unimodular group of V, the kernel of the homomorphism det$:\mathbf{G}\mathbf{L}(V)\rightarrow D^*_{ab}$. Choosing a basis of V allows one to identify this group with $\mathbf{S}\mathbf{L}_n(D)$.

We say that an automorphism $u$ of V is a transvection if there exist a vector $v$ of V and a linear form $\varphi$ on V such that $\varphi (v) = 0$ and $u(x) =x+v\varphi (x)$ for every $x$ in V. When $n\geqslant 2,u$ is a transvection if and only if there exists a basis of V with respect to which the matrix of $u$ is of the form $B_{ij}(\lambda )$. Theorem 1 implies the following corollary.