Partial Projective Planes
Partial Projective Planes
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Suppose some line g has type t^. Let P,Qeg have valence 2 and let Reg have valence 3. Only one point off g is joined to both P and Q. The two lines other than g through R each contain a point of valence 6 off g. By Proposition 2.3, these two points of valence 6 are joined to both P and Q, a contradiction. Hence no line has type t.. A similar argu- ment shows that no line has type t.,. Letting b. denote the number of lines of type t . , we now obtain b, + b„ = 20, b 1 + 2b 2 = 6vg, and b-j^ + 3b... 2 = 3v 3# The second and third of these equations imply that b, and b 2 are divisible by 3, a conclusion which contradicts the first equation.
Corollary 5.16 Any ppp of order 5 with b = 20 is extend- ible.
Theorem 5 .17 We have B Q (5) = {7, 19, 31} US, where S £ {8, 9,. . ., 18}.
Proof. In figure 5i a set of mopls of order 5 is shown.
The corresponding ppp ft has a line of type (6,6,6,2,2,2), so b = 19. One may check that ft is maximal by hand, or ar- gue as follows. By Theorem 3.13, Proposition 5.14, and Corollaries 5.13, 5.16 every ppp of order 5 with 20
What to read after Partial Projective Planes?
You can find similar books in the "Read Also" column, or choose other free books by Dow, Stephen John to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
Suppose some line g has type t^. Let P,Qeg have valence 2 and let Reg have valence 3. Only one point off g is joined to both P and Q. The two lines other than g through R each contain a point of valence 6 off g. By Proposition 2.3, these two points of valence 6 are joined to both P and Q, a contradiction. Hence no line has type t.. A similar argu- ment shows that no line has type t.,. Letting b. denote the number of lines of type t . , we now obtain b, + b„ = 20, b 1 + 2b 2 = 6vg, and b-j^ + 3b... 2 = 3v 3# The second and third of these equations imply that b, and b 2 are divisible by 3, a conclusion which contradicts the first equation.
Corollary 5.16 Any ppp of order 5 with b = 20 is extend- ible.
Theorem 5 .17 We have B Q (5) = {7, 19, 31} US, where S £ {8, 9,. . ., 18}.
Proof. In figure 5i a set of mopls of order 5 is shown.
The corresponding ppp ft has a line of type (6,6,6,2,2,2), so b = 19. One may check that ft is maximal by hand, or ar- gue as follows. By Theorem 3.13, Proposition 5.14, and Corollaries 5.13, 5.16 every ppp of order 5 with 20
What to read after Partial Projective Planes?
You can find similar books in the "Read Also" column, or choose other free books by Dow, Stephen John to read onlineMoreLess
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