# Any abelian normal subgroup normalizes an abelian subgroup of maximum order

## Statement

### Existential version

In symbols: let $P$ be a group of prime power order and $B$ be an abelian normal subgroup of $P$. Then, there exists an abelian subgroup of maximum order $A$ in $P$ such that $B \le N_P(A)$.

### Replacement version

let $P$ be a group of prime power order and $B$ be an abelian normal subgroup of $P$. Let $A$ be any abelian subgroup of maximum order in $P$. Then, there exists an abelian subgroup of maximum order $A_\infty$ in $P$ such that $B \le N_P(A_\infty)$ and such that $A_\infty$ is contained in the subgroup $\langle A^B \rangle$, i.e., the closure of $A$ under the action of $B$ by conjugation.

## Related facts

Note that this fact is uninteresting for small orders for the following silly reason: for small orders, it is also true that, among the abelian subgroups of maximum order, there exists a normal subgroup. The existence of a normal subgroup that is abelian of maximum order is obviously substantially stronger than this statement.

Below, we indicate which of a bunch of stronger statements is true, and why. In particular, we note from this that the smallest examples of interest where the stronger statements do not hold are $2^9 = 512$ for the prime 2 and $p^{10}$ (or possibly higher for some $p$, depending on other replacement results that are dependent on the value of $p$ -- see Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one for instance) for odd primes.

Value $n$ so we're considering groups of order $p^n$ Is every abelian subgroup of maximum order normal? Does there always exist an abelian subgroup of maximum order which is normal? Explanation
1 Yes Yes All groups of order $p$ are abelian, so the whole group is the unique abelian subgroup of maximum order.
2 Yes Yes All groups of order $p^2$ are abelian, so the whole group is the unique abelian subgroup of maximum order.
3 Yes Yes If the group is abelian, it is itself the unique subgroup of maximum order, which is normal. Otherwise, there exist subgroups of order $p^2$, all of which are abelian and normal.
4 Yes Yes If the group is abelian, it is itself the unique subgroup of maximum order, which is normal. Otherwise, by the existence of abelian normal subgroups of small prime power order, there exist abelian subgroups of order $p^3$ and index $p$. Using the fact that subgroup of index equal to least prime divisor of group order is normal (or instead that nilpotent implies every maximal subgroup is normal) we conclude that all these subgroups are normal.
5 No (?) Yes If the maximum order for an abelian subgroup is $p^4$ or $p^5$, then every subgroup of that order is normal. If the maximum order is $p^3$ or smaller, we can use the existence of abelian normal subgroups of small prime power order to conclude that there exists an abelian normal subgroup of the same order.
6 No (?) Yes If the maximum order for an abelian subgroup is $p^5$ or $p^6$, we can use that every subgroup of that order is normal. If the maximum order is $p^4$, use abelian-to-normal replacement theorem for prime-square index. If the maximum order is $p^3$ or less, use existence of abelian normal subgroups of small prime power order.
7 No Yes If the maximum order for an abelian subgroup is $p^6$ or $p^7$, we can use that every subgroup of that order is normal. If the maximum order is $p^5$, use abelian-to-normal replacement theorem for prime-square index. If the maximum order is $p^4$ or less (actually, it cannot be less), use existence of abelian normal subgroups of small prime power order.
8 No Yes Order $p^7$ is taken care of because prime power order implies every maximal subgroup is normal. For an odd prime $p$, the abelian-to-normal replacement theorem for prime-cube index for odd prime as well as abelian-to-normal replacement for prime-square index show that if there exists an abelian subgroup of order $p^5$ or $p^6$, there is an abelian normal subgroup of the same order. Further, the existence of abelian normal subgroups of small prime power order takes care of order up to $p^4$.
Note: The statement is also true for $p = 2$ but the $2^5$ in $2^8$ case needs to be proved more indirectly and cannot directly use the replacement theorem for prime-cube index.
9 No No for $p = 2$
Yes for odd primes
Order $p^8$ is taken care of because prime power order implies every maximal subgroup is normal. For an odd prime $p$, the abelian-to-normal replacement theorem for prime-cube index for odd prime as well as [[abelian-to-normal replacement for prime-square index show that if there exists an abelian subgroup of order $p^6$ or $p^7$, there is an abelian normal subgroup of the same order. Finally, the Jonah-Konvisser congruence condition on number of abelian subgroups of small prime power order for odd prime handles orders up to $p^5$.
For the prime $p = 2$, see Abelian-to-normal replacement fails for prime-cube index for prime equal to two. The smallest example for that is a subgroup of order $p^6$ in a group of order $p^9$.

## Facts used

1. Stable version of Thompson's replacement theorem for abelian subgroups, which in turn follows from Thompson's replacement theorem for abelian subgroups

## Proof

### Proof of replacement version

The proof follows directly from Fact (1), and the observation that the hypotheses apply because, on account of $B$ being abelian normal, any abelian subgroup of maximum order must normalize it.

### Proof of existential version

This is weaker than, and hence follows from, the replacement version.