Is it possible that an element of a sylow basis of a finite solvable group G lies in the frattini subgroup of G?(i.e could an element of a sylow basis of a finite solvable group G be a non-generator subgroup?)
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You should tell the reader what a Sylow basis is. Assuming that it s a set of pairwise permutable Sylow subgroups of $G,$ one for each prime divisor, the answer is no. For if $P_{1}$ is a Sylow $p_{1}$-subgroup of $G$ belonging to a basis, then deleting $P_{1}$from the generating set given by the Sylow basis leaves a set which generates a Hall $p_{1}^{\prime}$-subgroup of $G$- in particular, a proper subgroup of $G.$
answered Nov 10, 2012 at 21:05
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$\begingroup$ It is the exact definition of sylow basis of a finite group G. thank you so much. $\endgroup$ Commented Nov 10, 2012 at 22:24