# 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 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

## 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.