By Borovik A. V.

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**Example text**

So we may assume further that U1 ,U2 ≤ P⊥ . In particular, P ≤ P⊥ . Next we show that we may assume that U1 = U2 = P⊥ . Suppose first that U1 = U2 . If Ui = H, ui for i = 1, 2, let W0 be a complement for U1 + U2 in P⊥ , and W = W0 , u1 + u2 ; then h can be extended by the identity on W to an isometry on P⊥ . If U1 = U2 , take any complement W to U1 in P⊥ . In either case, the extension is an isometry of P⊥ which acts as the identity on a hyperplane H of P⊥ containing H. So we may replace U1 ,U2 , H by P⊥ , P⊥ , H .

Hence, again by the FTPG, the automorphism group of the geometry is induced by the group of semilinear transformations of V which preserve the set of pairs {(x, y) : B(x, y) = 0}. These transformations are composites of linear transformations preserving B up to a scalar factor, and field automorphisms. It follwos that, if F = GF(2), the automorphism group of the geometry has a unique minimal normal subgroup, which is isomorphic to PSp(4, F). 56 5 Unitary groups In this section we analyse the unitary groups in a similar way to the treatment of the symplectic groups in the last section.

In either case, the extension is an isometry of P⊥ which acts as the identity on a hyperplane H of P⊥ containing H. So we may replace U1 ,U2 , H by P⊥ , P⊥ , H . Let P = x and let x = uh − u for some u ∈ U1 . We have B(x, x) = 0. In the orthogonal case, we have q(x) = q(uh − u) = q(uh) + q(u) − B(uh, u) = 2q(u) − B(u, u) = 0. ) So P is flat, and there is a hyperbolic plane u, v , with v ∈ / P⊥ . Our job is to extend h to the vector v. To achieve this, we show first that there is a vector v such that uh, v ⊥ = u, v ⊥ .