Existence of abelian normal subgroups of small prime power order

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 fact about::abelian normal subgroup (hence, an fact about::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 $$p^k$$.

Similar facts

 * Existence of abelian ideals of small prime power order in nilpotent Lie ring

Opposite facts

 * Alperin's theorem on non-existence of abelian normal subgroups of large prime power order for odd prime
 * Alperin's theorem on non-existence of abelian normal subgroups of large prime power order for prime equal to two

Corollaries

 * Abelian-to-normal replacement theorem for prime-cube order
 * Abelian-to-normal replacement theorem for prime-fourth order (actually, this is a corollary of the stronger version group of prime-sixth or higher order contains abelian normal subgroup of prime-fourth order for prime equal to two)

Other related facts

 * Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime: In particular, this guarantees that for odd $$p$$, and $$0 \le k \le 5$$, the existence of an abelian subgroup of order $$p^k$$ implies the existence of an abelian normal subgroup of order $$p^k$$.
 * Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime
 * Congruence condition on number of abelian subgroups of small prime power order and bounded exponent for odd prime
 * Lower bound on order of maximal among abelian normal subgroups in terms of order of finite p-group

Facts used

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