Congruence condition on number of abelian subgroups of prime index

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 $$p + 1$$.

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) uses::Jonah-Konvisser line lemma
 * 2) uses::Cyclic over central implies abelian

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.