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By Penrose R.
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Additional resources for A lecture on 5-fold symmetry and tilings of the plane
Example text
Y : i E Z) = NY. 1 1 0 Thus N. is an r-group normalized by (x). 1 Choose Z maximal such that N = (Q. , ... , Q. ) is an r-group normalized by (x). 1 l Z 1 Assume for simplicity that N = (Q1' ... , QZ)' Of course x E NG(N). Let T be a Sylow r-subgroup of G containing' N. if N;;J T or there exists if- By 9. I, either N such that NZ ~ NG(N), z ET. If N
Z(R)y E Q. and so N. y : i E Z) = NY. 1 1 0 Thus N. is an r-group normalized by (x). 1 Choose Z maximal such that N = (Q. , ... , Q. ) is an r-group normalized by (x). 1 l Z 1 Assume for simplicity that N = (Q1' ... , QZ)' Of course x E NG(N). Let T be a Sylow r-subgroup of G containing' N. if N;;J T or there exists if- By 9. I, either N such that NZ ~ NG(N), z ET. If N
Proof. Let G be a minimal counter example. obViously a minimal simple group of odd order. Then G is Let P be a Sylow p- subgroup, Q a Sylow q-subgroup of G. (i) V1 (R, r')::::: {1 ) for any Sylow r-subgroup of G. G For if X E V1 (R, r'), there exists a Sylow r'-subgroup R' G ::J X. Then XG ::::: ~R' ::::: ~' and X G C R' is a proper normal subgroup of G. Let H be a maximal subgroup of G. (ii) If 11T(F(H)) I : : : 2, then F(H) is non-cyclic. For if r::::: max (P, q), then H::::: CH(F(H)r') clearly and CG(F(H)r) ::::: CG(F(H)) ~ F(H), assuming F(H) is cyclic.