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Let $ G $ is a finite group. Suppose $ H \unlhd G $ such that $ G/H $ is supersoluble. Suppose $ H \cap M $ is either $ H $ or a maximal subgroup of $ H $ for any maximal subgroup of $ G $. Suppose that $ P_{H}(M) = \langle h \in H \mid \langle h \rangle M = M \langle h \rangle \rangle = H $ for any maximal subgroup $ M $ of $ G $. Let $ p \in \pi(G) $ be an odd prime, then $ G $ is $ p $-supersoluble.

For proof: Since $ P_{H}(M) = H $, then $ H $ is soluble, then $ G $ is soluble. Now assume $ G $ is not $ p $-supersoluble and $ G $ is a counterexample of smallest order. Let $ N $ be a minimal normal subgroup of $ G $, then $ N $ is an elementary abelian $ q $-group for prime $ q $. I show $ G/N $ is $ p $-supersoluble group and $ N $ is unique minimal normal subgroup. Now $ \Phi(G) = 1 $ or $ \Phi(G) \neq 1 $?

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