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 $n\geq 1+k(k-1)/2$ (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

• 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 $0\leq k\leq 5$, 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. 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 $p^{n}$. $n\geq 1+k(k-1)/2$.

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 $l(l+1)/2\geq n$: This follows from fact (1).
3. $l\geq k$: Since $l(l+1)/2\geq n\geq k(k-1)/2+1$, $l$ is strictly greater than $k-1$, so $l\geq k$.
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

• Paper:Burnside12^{More info}
• Large abelian subgroups of p-groups by Jonathan Lazare Alperin, Transactions of the American Mathematical Society, Volume 117, Page 10 - 20(Year 1965): ^{Official copyMore info}