Jonah-Konvisser congruence condition on number of elementary abelian subgroups of small prime power order for odd prime

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This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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Statement

Statement in terms of a universal congruence condition

Let $p$ be an odd prime. Suppose $0\leq k\leq 5$.

Let ${\mathcal {S}}$ be the one-element set comprising an elementary abelian subgroup of order $p^{k}$. Then, ${\mathcal {S}}$ is a Collection of groups satisfying a universal congruence condition (?) for the prime $p$. In particular, ${\mathcal {S}}$ is a Collection of groups satisfying a strong normal replacement condition (?) for $p$ and hence also a Collection of groups satisfying a weak normal replacement condition (?) for $p$.

Hands-on statement

Suppose $p$ is an odd prime number and $0\leq k\leq 5$. Suppose $G$ is a finite $p$-group having an elementary abelian subgroup of order $p^{k}$.

The statement has the following equivalent forms:

1. The number of elementary abelian subgroups of $G$ of order $p^{k}$ is congruent to $1$ modulo $p$.
2. The number of elementary abelian normal subgroups of $G$ of order $p^{k}$ is congruent to $1$ modulo $p$.
3. If $G$ is a subgroup of a finite $p$-group $L$, then the number of elementary abelian subgroups of $G$ of order $p^{k}$ that are normal in $L$ is congruent to $1$ modulo $p$.

In particular, if $G$ has an elementary abelian subgroup of order $p^{k}$, then $G$ has an elementary abelian normal subgroup of order $p^{k}$. In fact, $G$ has an elementary abelian p-core-automorphism-invariant subgroup of order $p^{k}$, and the number of elementary abelian $p$-core-automorphism-invariant subgroups of $G$ of order $p^{k}$ is also congruent to $1$ modulo $p$.

Corollary in terms of normal rank

In particular, this shows that for $p$ an odd prime and $G$ a $p$-group:

• If the rank of $G$ is less than or equal to $5$, the normal rank of $G$ is equal to the rank.
• If the normal rank is at most $4$, the rank equals the normal rank.

Related facts

Similar general facts

• Congruence condition on number of subgroups of given prime power order
• Congruence condition relating number of subgroups in maximal subgroups and number of subgroups in the whole group

Similar replacement theorems

• Elementary abelian-to-normal replacement theorem for prime-square order is a weaker version proved much more easily using the same techniques.
• Jonah-Konvisser abelian-to-normal replacement theorem

For a full list of replacement theorems (including many of a completely different flavor) refer Category:Replacement theorems.

Opposite facts

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References

Journal references

• Counting abelian subgroups of p-groups: a projective approach by Marc Konvisser and David Jonah, Journal of Algebra, ISSN 00218693, Volume 34, Page 309 - 330(Year 1975): ^{PDF (ScienceDirect)More info}