Congruence condition on number of abelian subgroups of prime index

This article is about a congruence condition.
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Definition

Suppose $p$ is a prime number and $P$ is a finite $p$-group that has an abelian maximal subgroup, i.e., an abelian subgroup of index $p$. Then, the number of abelian subgroups of index $p$ is congruent to $1$ modulo $p$.

More strongly, if $P$ is non-abelian, the number is either $1$ or $p+1$.

Related facts

Similar facts for groups

• Elementary abelian-to-normal replacement theorem for prime-square order
• Elementary abelian-to-normal replacement theorem for prime-cube and prime-fourth order
• Jonah-Konvisser elementary abelian-to-normal replacement theorem
• Jonah-Konvisser abelian-to-normal replacement theorem
• Jonah-Konvisser congruence condition for abelian subgroups of prime-square index

Similar facts for rings

• Congruence condition on number of abelian subrings of prime index in nilpotent ring

Facts used

1. Jonah-Konvisser line lemma
2. Cyclic over central implies abelian

Proof

Proof using the line lemma (weaker version)

Given: A prime $p$, a finite $p$-group $P$. An abelian maximal subgroup $M$ of $P$. In particular, the index of $M$ in $P$ equals $p$.

To prove: The number of abelian maximal subgroups of $P$ is congruent to $1$ modulo $p$.

Proof:

1. If $N$ is another abelian maximal subgroup of $P$, and maximal subgroup containing $M\cap N$ is abelian: Since both $M$ and $N$ are abelian and maximal, $MN=P$, so $M\cap N$ is in the center of $P$. In particular, any subgroup generated by $M\cap N$ and one element is abelian. Since $M\cap N$ has index $p^{2}$, we obtain that any maximal subgroup containing it is generated by it and one element, hence is abelian. (see fact (2)).
2. If $N$ is also an abelian maximal subgroup of $P$, the number of abelian maximal subgroups containing $M\cap N$ is congruent to $1$ modulo $p$: From the previous step, the number of such subgroups is equal to the number of subgroups of index $p$ containing $M\cap N$, which in turn equals the number of subgroups of $P/(M\cap N)$ of order $p$, which is $p+1$.
3. The result now follows from fact (1), since we have essentially shown that $M$ is an origin.

Proof of stronger version for non-abelian groups

Given: A prime $p$, a finite non-abelian $p$-group $P$. An abelian maximal subgroup $M$ of $P$. In particular, the index of $M$ in $P$ equals $p$.

To prove: The number of abelian maximal subgroups of $P$ is $1$ or $p+1$.

Proof: We consider three cases:

1. There is exactly one abelian maximal subgroup $M$: In this case, we are done.
2. There are two abelian maximal subgroups $M,N$, and all abelian maximal subgroups contain $M\cap N$: First, we note that any maximal subgroup containing $M\cap N$ is generated by $M\cap N$ and a single element. Since $MN=P$, $M\cap N$ is central, so each such maximal subgroup containing it is abelian. Thus, the abelian maximal subgroups are in bijection with subgroups of order $p$ in $P/(M\cap N)$.
3. There are three distinct abelian maximal subgroups $M,N,Q$ such that none contains the intersection of the other two: Since $MN=P$, $M\cap N$ is in the center of $P$. Similarly, $M\cap Q$ is in the center of $P$ and $N\cap Q$ is in the center of $P$. Thus, we have three distinct subgroups of index $p^{2}$ in the center of $P$. The join is either the whole group or has index $p$. If it is the whole group, we are done. If it has index $p$, we apply fact (2) and we are again done.

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}