Projective planes of order 12 do not have a non-abelian group of order 6 as a collineation group.
Projective planes of order 12 do not have a non-abelian group of order 6 as a collineation group.
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12 阶射影平面没有 6 阶非阿贝尔群作为直射群。
DOI：
10.1515/crll.1981.326.152
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发表时间：
1981
期刊：
影响因子：
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通讯作者：
Z. Janko
中科院分区：
文献类型：
--
作者：
T. Trung;Z. Janko
定理A的证明。设P是一个12阶射影平面，它有一个与π 3同构的直射群，其中ρ是一个3阶直射群，τ是一个对合。我们有ρ = τ~ ρτ = ρ~。直射投影ρ有r个不动点（r = 1、4、7、10或13），它们都位于由ρ固定的直线o上（因为ρ的固定结构显然不可能是3阶子平面），并且由ρ固定的r条直线通过由ρ固定的点O，其中O e o。我们看到τ固定O和o，所以O是τ的中心，τ的轴o'穿过O。因为ρ固定0，所以O也是τ的中心。因此，τ和τ都固定通过O的所有行，因此也固定通过O的所有行。由此可知，ρ是一个以o为轴，以O为中心的π。因为的阶不是素数幂，所以O' = O。
The proof of theorem A. Let P be a projective plane of order 12 which possesses a collineation group isomorphic to Σ3, where ρ is a collineation of order 3 and τ is an involution. We have ρ = τ~ ρτ = ρ~. The collineation ρ has exactly r fixed points (r = l, 4, 7, 10 or 13) which all lie on a line o fixed by ρ (since the fixed structure of ρ obviously cannot be a subplane of order 3) and the r fixed lines by ρ pass through a point O, fixed by ρ where O e o. We see that τ fixes O and o and so O is the center of τ and the axis o' of τ passes through O. Since ρ fixes 0, it follows that O is also the center of τ. Hence both τ and τ fix all lines through O and so also = fixes all lines through O. It follows that ρ is an elation with axis o and center O. Since the order of is not a prime power, it follows that o' = o.