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Prove that a group of order 56 has a normal Sylow p-subgroup for some prime divding its order.

56=2^{3}\cdot 7\,

The number of Sylow 2-subgroups is either 1 or 7.

The number of Sylow 7-subgroups is either 1 or 8.

Suppose there are 8 Sylow 7-subgroups which account for 8\cdot (7-1)=48\, elements without the identity, and two Sylow 2-subgroups, the maximum order of intersection of which is 4, including identity, which accounts for 12 elements. But 48+12=60\,, which is a contradiction of the group order 56.

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