Symmetry And Group

## 2-Signalizers of Finite Simple Groups by Kondratiev A. S., Mazurov V. D.

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By Kondratiev A. S., Mazurov V. D.

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Such that Fa + F}:;a is a regular plane. (3) We can also show that we have a E V, }:; E G such that Fa + F}:;a is a degenerate plane. Pick E G with =I' ± 1v' Then there is an a in V such that Fa =I' Fa. If q(a, a) = 0 we are done, so assume q(a, a) =I' O. Pick b E V - (Fa + Fa) with q(b, a) =0 and q(a, b) = q(a, a) =I' O. O. 42 T. O'MEARA By Witt's Theorem there is a T in SPn( V) with Ta =

Oa» will do the job. The fixed space of this Ta -I is equal to P + Fa (which :J P). PROOF. First let us suppose that we have a T with res 'Ta < res a. Let L ~ H be the residual and fixed spaces of T. Of course H = L *. Put a l = 'Ta. Then the equation a = T -Ial implies that L g R I' hence H ;Q PI' hence there is an a in V with ala = a and Ta =1= a. Then q(a, aa) = q(a, T-laJa) = q(a. 1. Conversely, let there be an a in V with q(a, aa) =1= O. Put T = -I T, . 3. Hence 'Taa = a. Now the residual space of Ta is contained in F(aa - a) + R which is R since aa - a E R; hence the fixed space of Ta contains P; but a is in the fixed space of Ta though not in the fixed space P of a; hence the fixed space of Ta contains P + Fa which strictly contains P.

Then there is an X in V with (JX =I=- x. Let C and C be the two configurations containing x. Then (JX does not belong to one of them, say (JX fl C. So (JC =I=- C. So =I=- I. In other words, the kernel is trivial and we have an injective homomorphism Spi V) >- CS 6 · But Spi V) has 6! 2. So Spi V) is isomorphic to CS 6 , as required. D. 2. Centers Note that PSPn(V) is not commutative. 8. So SPn( V) is not commutative too. 1. PSPn( V) is centerless and cen SPn( V) = (± 1v). 38 O. T. O'MEARA PROOF.

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