Existence of abelian normal subgroups of small prime power order

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History

This result appears to have been first noted by Burnside, in a paper published in 1912.

Statement

Suppose p is a prime number and G is a finite p-group of order p^n (i.e., a group of prime power order). Then, if k is a nonnegative integer such that n \ge 1 + k(k-1)/2 (i.e., n > k(k-1)/2), G has an Abelian normal subgroup (?) (hence, an Abelian normal subgroup of group of prime power order (?)) of order p^k.

Particular cases

This table lists, for various values of k, the smallest value of n guaranteed by the theorem and the actual smallest value of n such that any group of order p^n contains an abelian normal subgroup of order p^k.

k Theorem-guaranteed n Actual smallest n Actual smallest n for p = 2
1 1 1 1
2 2 2 2
3 4 4 4
4 7 6 or 7 6

Related facts

Similar facts

Opposite facts

Corollaries

Other related facts

Facts used

  1. Lower bound on order of maximal among abelian normal subgroups in terms of order of finite p-group
  2. Finite nilpotent implies every normal subgroup contains normal subgroups of all orders dividing its order
  3. Abelianness is subgroup-closed

Proof

Given: A finite p-group G of order p^n. n \ge 1 + k(k-1)/2.

To prove: G has an abelian normal subgroup of order p^k.

Proof:

  1. Let H be a subgroup of G that is maximal among abelian normal subgroups: Note that such a subgroup clearly exists.
  2. If the order of H is p^l, we have l(l + 1)/2 \ge n: This follows from fact (1).
  3. l \ge k: Since l(l+1)/2 \ge n \ge k(k-1)/2 + 1, l is strictly greater than k - 1, so l \ge k.
  4. We now use Fact (2) to conclude that H contains a subgroup of order p^k that is normal in G, and Fact (3) to conclude that this subgroup is abelian.

References

Journal references