THE GEOMETRY OF GL ( 2 , q ) IN TRANSLATION PLANES OF EVEN ORDER q 2

In this article we show the following: Let II be a transla2 tion plane of even order q that admits GL(2,q) as a collineation group. Then u is either Desarguesian, Hall or Ott-Schaeffer.


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N. L. JOHNSON or 0tt- Schaeffer. In [9], it is shown that if the assumption on the kernel is dropped, the planes admitting SL(2,q) must have properties quite similar to the Desarguesian, Hall or 0tt-Schaeffer planes but it is an open question whether such planes must, in fact, be in one of these three classes.
In the dimension 2 situation, if the plane admits SL(2,2s) then the plane also admits GL(2,2s) (see (2,2)). So, in particular, the Hall and Ott-2 Schaeffer planes of even order q must admit GL(2,q) as a collineation group.
In this paper we observe that this situation can be reversed (see Theorem (.)).
We assume the reader is familiar with the papers [9] and [i0]. 2. THE MAIN THEOREM.
Since the kern homology group K of order 2s-i comnmtes with elements of the linear translation complement and SL(2,2s) is linear by [10] (see the proof of (2.1)), we have a group SL(2,2 s) K (since SL(2,2s) is simple).
If a translation plane is of dim 2 then the group in the linear translation complement generated by all Baer involutions is always SL (2,2s) if there are no elations and the group is nonsolvable (see [i0], (3.27)). Let Qx,Q, respectively fix the Baer subplanes l,W2,w 3 pointwise.
Thus, (QX) nQ (i> for x-(Q). (Theorem (5.1)) and thus (2.h) applies 2 (2.6) THEOREM. Let u be a translation plane of even order q which admits GL(2,q) where the sylow 2-groups fix subsets of order q that are contained in components. Then is an Ott-Schaeffer plane.
Let G be in the center of GL(2,q). Then G must fix each Baer subplane which is fixed pointwise by an involution and must fix each line of the orbit of length q+l.
The q-1 Baer subplanes fixed pointwise by elements of a sylow 2-group of SL(2,q) cover the components other than the components in the orbit of length  (G) . Thus, l(Gi>l<2s-1 since @ is fixed-point-free and the -orbit of has length < 2s+l since fixes X and X has 2 s+l components So, q-1 < (s2-1) (2S+l) 2s-1 K q-i so that X has order and the orbit length of i is +i. Let the fixed point subplane of (Gg> be n. Let T be an involution that maps X Y. Thus u and uT both contain X and Y. Since XNY O, =UT (X and Y are r-dim subspaces of n and nT).
Since (o-lg) also fixes pointwise, then (-ig)(g)=g2 fixes Note that w contains precisely X and Y, for otherwise w would contain all q+l subplanes of order as is regular on remaining subplanes # X or Y.
We assert that XI,X 2 and X I,Xj' J 2, share only the components of X I PROOF. If the subplanes share a component then (,XI> is a Baer subplane (smallest subplane properly containing X I has order ()2) so X I X I ,X 2=WXI X J #2' but then X I,X2,Xj are in ,X 2 contrary to the ,j above.
We also assert that XI,X 2 and UX3,X 4 share no common component.
That is, either = or and every prime 2-primitive divisor element fixes pointwise. Since SL(2,q) is simple and fixes , either SL(2,q) fixes J pointwise or is faithful on . The first alternative cannot exist by [9] (4.2). So SL(2,q) is faithful on and thus =.
If is f.p.f, on y= 0 then since SL(2,q) fixes y= 0, we may apply or previous argument of Case i. That is, a Sylow 2-subgroup Q of SL(2,q) must fix a subspace X of pointwise. Since is f.p.f, on y= 0 and fixes , then II q contrary to our assumptions.