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Note that this fact is superseded by the combination of the following two facts: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