Elementary abelian-to-normal replacement theorem for large primes

Hands-on statement
Let $$P$$ be a group of prime power order, say $$p^r$$, and let $$A$$ be an elementary abelian subgroup of $$P$$ of order $$p^k$$.

Suppose $$k \le (p + 5)/4$$, and $$p$$ is odd.

Then, there exists an elementary abelian normal subgroup $$B$$ of $$P$$ satisfying:


 * $$B$$ has order $$p^k$$ (same as $$A$$)
 * $$B$$ is contained in the normal closure of $$A$$

Statement in terms of normal replacement conditions
Suppose $$k$$ is a natural number and $$p \ge 4k - 5$$ is an odd prime. Then, the singleton collection of the elementary abelian group of order $$p^k$$ is a fact about::collection of groups satisfying a strong normal replacement condition. In particular, it is also a fact about::collection of groups satisfying a weak normal replacement condition.

Statement/corollary in terms of normal rank

 * For an odd prime $$p$$, a $$p$$-group whose rank is at most $$(p + 5)/4$$ has the property that its rank equals its normal rank.
 * For an odd prime $$p$$, a $$p$$-group whose normal rank is at most $$(p + 1)/4$$ has the property that its rank equals its normal rank.

Related facts
Note that this fact is superseded by the combination of the following two facts:


 * Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one
 * Jonah-Konvisser abelian-to-normal replacement theorem

Other related facts

 * Elementary abelian-to-2-subnormal replacement theorem

Main fact used

 * 1) uses::Abelian-to-normal replacement theorem for prime exponent, which is Theorem B of the same paper. It states the following: Suppose $$P$$ is a finite fact about::group of prime exponent: group of prime power order, say $$p^r$$, and with exponent $$p$$ (so every element has order $$p$$). Suppose $$A$$ is an abelian subgroup of order $$p^n$$, and nilpotency class at most $$p + 1$$.

Then, there exists an fact about::abelian normal subgroup $$B$$ of $$P$$ such that:


 * $$B$$ is contained in the normal closure of $$A$$ in $$P$$
 * $$B$$ has the same order (i.e., $$p^n$$) as $$A$$

Proof
The proof given here is (largely) the same as that presented in the original paper where the theorem appeared.

Overall proof strategy
We fix the prime $$p$$ beforehand. The overall proof strategy is to use a double induction, first on $$k$$ and then on $$r$$. Note that the variables $$k$$ and $$r$$ differ in one important respect: $$k$$ is bounded from above in terms of $$p$$, so that induction proceeds only for finitely many steps for any fixed $$p$$ (though the number of steps increases as $$p$$ increases), whereas $$r$$ is not bounded.

By double induction, we mean that in order to prove the statement for a particular pair $$(k,r)$$, we assume the statement to be true for $$(l,r)$$ where $$0 \le l < k$$ and and for $$(k,s)$$ where $$0 \le s < r$$.

We use some simple reasoning to ultimately reduce to the case of Fact (1), which is Theorem A of the same paper.

Base case for induction
This case is obvious.

Inductive step
Given: A group $$P$$ of order $$p^r$$, $$p$$ prime. An elementary abelian subgroup $$A$$ of $$P$$ of order $$p^k$$, with $$k \le (p + 5)/4$$.

To prove: There exists a normal subgroup $$B$$ of $$P$$ of order $$p^k$$ such that $$B$$ is contained in the normal closure of $$A$$ in $$P$$.

Part one: reduction to the case that $$A$$ is 2-subnormal and its normal closure has exponent $$p$$
This part uses the inductive hypothesis on $$r$$.

To prove: It suffices to consider the case where $$A$$ is 2-subnormal and the normal closure of $$A$$ in $$P$$ has exponent $$p$$ and nilpotency class at most $$k$$, which is less than $$p$$.

Proof: We will show that there is a subgroup $$A_1$$ lying in the normal closure of $$A$$ such that $$A_1$$ is 2-subnormal and its normal closure has exponent $$p$$ and nilpotency class at most $$k$$, which is less than $$p$$. In particular, $$A_1^P \le A^P$$ Thus, if we show that the result holds for all groups satisfying the conditions established for $$A_1$$, we get the result for $$A_1$$ and hence for $$A$$.

Steps (PA4), (PA5), and (PA6) establish the existence of $$A_1$$ contained in $$A^P$$ satisfying all the required properties.

Part two: reduction to the case where there is a self-centralizing (in $$A^P$$) elementary abelian normal (in $$P$$) subgroup of order $$p^{k-1}$$
This part uses the inductive hypothesis on $$k$$. Note that we assume that $$A$$ satisfies the properties established for $$A_1$$ in part one.

To prove: It suffices to consider the case where $$A$$ is 2-subnormal and there is an elementary abelian normal subgroup $$B$$ of $$P$$ of order $$p^{k-1}$$ and inside $$A^P$$ such that $$C_{A^P}(B) = B$$.

Proof:

Note that the lengthy argument in (PB3) is similar to the proof that maximal among abelian normal implies self-centralizing in nilpotent. The main difference is that here, normality is in $$P$$ but the self-centralizing subgroup condition is in the smaller subgroup $$A^P$$. Thus, although the proof technique remains the same, we cannot directly use the other statement as a black box.

Similar arguments to (PB3) are also used in the proof of Thompson's replacement theorem for abelian subgroups and Thompson's critical subgroup theorem.