Elementary abelian-to-normal replacement theorem for large primes

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Statement

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 Collection of groups satisfying a strong normal replacement condition (?). In particular, it is also a 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Note that this fact is superseded by the combination of the following two facts:

Facts used

Main fact used

1. 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 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 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 $A$

Other facts used

No. Assertion
O1 prime power order implies nilpotent, nilpotent implies every maximal subgroup is normal. The upshot is that for a finite $p$-group, any maximal subgroup is normal and has index $p$.
O2 prime-base logarithm of order of 2-subnormal subgroup of group of prime power order gives upper bound on nilpotency class of its normal closure
O3 ‎p-group of nilpotency class less than p implies exponent is maximum of orders over any generating set
O4 cyclic over central implies abelian
O5 prime power order implies center is normality-large
O6 normality is quotient-transitive
O7 normality is centralizer-closed
O8 normality satisfies image condition, normality satisfies inverse image condition

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. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

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 $A$.

Step no. Assertion Facts used Given data/previous part steps used Steps used Explanation
PA1 If $A = P$, we are done. Hence, we can restrict henceforth to the case $k < r$, so $A$ is proper in $P$. $P$ is normal in itself, hence $A$ is normal and we are done.
PA2 There exists a maximal subgroup $Q$ of $P$ containing $A$. $Q$ is normal in $P$ and has order $p^{r - 1}$. In particular, $Q$ contains the normal closure $A^P$ of $A$ in $P$. (O1) Step (PA1): $A$ proper in $P$
PA3 There exists an elementary abelian normal subgroup $A_1$ of $A^P$ of order $p^k$. Inductive hypothesis on $r$. Step (PA2): $Q$ contains $A^P$. [SHOW MORE]

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:

Step no. Assertion Facts used Given data/previous part steps used Steps used Explanation
PB1 $A$ contains an elementary abelian subgroup $A_2$ of order $p^{k-1}$. An elementary abelian subgroup of order $p^k$ contains elementary abelian subgroups of all orders $p^l, 0 \le l \le k$. $A$ has order $p^k$.
PB2 There exists an elementary abelian normal subgroup $B$ of $P$ lying inside $A^P$ of order $p^{k-1}$. Inductive hypothesis on $k$ (PB1) [SHOW MORE]
PB3 If $C_{A^P}(B) \ne B$, then we can find an abelian normal subgroup $A_3$ of $P$ containing $B$ and of order $p^k$ such that $A_3 \le A^P$ (O1), (O4)-(O8) -- (PB2) [SHOW MORE]
PB4 The $A_3$ constructed in (PB3), subject to the condition that $C_{A^P}(B) \ne B$, is elementary abelian -- $A^P$ has exponent $p$ (PB3) <toggledisplay>Since $A^P$ has exponent $p$, the subgroup $A_3$ is abelian of exponent $p$, hence is elementary abelian.
PB5 It suffices to consider the case $C_{A^P}(B) = B$. -- -- (PB3), (PB4) By the previous two steps, if $C_{A^P}(B) \ne B$, then $A_3$ works as a replacement for $A$. Hence, it suffices to restrict attention to the case that $C_{A^P}(B) = B$.

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.