Exercise 12, § 10

Let E be a right vector space of dimension $\geqslant 1$ over a field K. Let $F(E)$ denote the normal subgroup of $\mathbf{GL}(E)$ consisting of the automorphisms $u$ such that the kernel of $1_E - u$ is of finite codimension (hence $F(E) = \mathbf{GL}(E)$ if E is finite-dimensional). Let $\mathbf{SL}(E)$ denote the normal subgroup of $\mathbf{GL}(E)$ generated by all the transvections (Exercise 11); it is contained in $F(E)$.

(a) If $\dim E \geqslant 2$, show that for every ordered pair of non-zero vectors $x, y$ of E, there exists a transvection or product of two transvections, which maps $x$ to $y$ (in other words $\mathbf{SL}(E)$ operates transitively on $E - \{0\}$).

(b) Let V, W be two hyperplanes of E, $\dot{x}_0 = x_0 + V$ a class mod. V distinct from V and $\dot{y}_0 = y_0 + W$ a class mod. W distinct from W. Show that if $\dim E \geqslant 2$, there exists a transvection or product of two transvections, which maps V to W and $\dot{x}_0$ to $\dot{y}_0$ (consider first the case where V and W are distinct).

(c) If $\dim E \geqslant 2$, show that any two transvections distinct from the identity are conjugate in the group $F(E)$.

(d) If $\dim E \geqslant 3$, show that any two transvections distinct from the identity are conjugate in the group $\mathbf{SL}(E)$ (reduce it with the aid of (b) to the case where the hyperplanes of the two transvections are identical, then use (a)).

(e) Suppose that dim E = 2. For any two transvections to be conjugate in $\mathbf{SL}(E)$, it is necessary and sufficient that the subgroup Q of K* generated by the squares of elements of K* be identical with K*. (If u is a transvection distinct from the identity, a a vector of E which is not invariant under u and $b = u(a) - a$, show that, for every transvection $u'$ conjugate to u in $\mathbf{SL}(E)$, $u'(a) - a = a\lambda + b\mu$, with $\mu = 0$ or $\mu \in Q$; for this use the fact that in a group G every product $sts^{-1}t$ is a product of squares.)

(f) Suppose that dim E $\geqslant 2$. For two dilatations $u, u'$ to be such that there exists a $v \in \mathbf{SL}(E)$ such that $u' = vuv^{-1}$, it is necessary and sufficient that the classes (Exercise 11) of u and $u'$ be the same (use (b)). Deduce that if the class of a dilatation is contained in the commutator group of K*, this dilatation belongs to the group $\mathbf{SL}(E)$ (observe that $(vuv^{-1})u^{-1} = v(uv^{-1}u^{-1})$).