Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup

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Statement

Suppose G is a Finite solvable group (?), and n is a natural number dividing the order of G. If there are exactly n elements of G whose n^{th} power is the identity element, then these elements form a subgroup. This subgroup is clearly a normal subgroup. In fact, it is a characteristic subgroup, and even better, is a fully characteristic subgroup.

Facts used

  1. Minimal normal implies elementary abelian in finite solvable
  2. Number of nth roots is a multiple of n

Proof

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References

Textbook references