Any abelian normal subgroup normalizes an abelian subgroup of maximum order
Contents
Statement
Existential version
Any abelian normal subgroup of a group of prime power order (i.e., any abelian normal subgroup of group of prime power order) normalizes an abelian subgroup of maximum order.
In symbols: let be a group of prime power order and be an abelian normal subgroup of . Then, there exists an abelian subgroup of maximum order in such that .
Replacement version
let be a group of prime power order and be an abelian normal subgroup of . Let be any abelian subgroup of maximum order in . Then, there exists an abelian subgroup of maximum order in such that and such that is contained in the subgroup , i.e., the closure of under the action of 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 for the prime 2 and (or possibly higher for some , depending on other replacement results that are dependent on the value of -- see Glauberman's abelian-to-normal replacement theorem for bounded exponent and half of prime plus one for instance) for odd primes.
Value so we're considering groups of order | 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 are abelian, so the whole group is the unique abelian subgroup of maximum order. |
2 | Yes | Yes | All groups of order 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 , 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 and index . 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 or , then every subgroup of that order is normal. If the maximum order is 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 or , we can use that every subgroup of that order is normal. If the maximum order is , use abelian-to-normal replacement theorem for prime-square index. If the maximum order is 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 or , we can use that every subgroup of that order is normal. If the maximum order is , use abelian-to-normal replacement theorem for prime-square index. If the maximum order is or less (actually, it cannot be less), use existence of abelian normal subgroups of small prime power order. |
8 | No | Yes | Order is taken care of because prime power order implies every maximal subgroup is normal. For an odd prime , 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 or , 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 . Note: The statement is also true for but the in case needs to be proved more indirectly and cannot directly use the replacement theorem for prime-cube index. |
9 | No | No for Yes for odd primes |
Order is taken care of because prime power order implies every maximal subgroup is normal. For an odd prime , 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 or , 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 . For the prime , 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 in a group of order . |
Facts used
- 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 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.
References
Textbook references
- Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 274, Theorem 2.6, Section 8.2 (Glauberman's theorem), ^{More info}