Omega-1 of odd-order class two p-group has prime exponent
This article states and (possibly) proves a fact that is true for odd-order p-groups: groups of prime power order where the underlying prime is odd. The statement is false, in general, for groups whose order is a power of two.
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Let be a p-group (not necessarily finite) of nilpotence class two, where is an odd prime. Then, the subgroup:
is a group of prime exponent: in other words, every non-identity element in has order . Equivalently, in a -group of nilpotence class two, the product of any two elements of order is either trivial, or has order .
- Formula for powers of product in group of class two
- Omega-j of odd-order class two p-group has exponent dividing p^j
Facts with similar proofs
Breakdown for the prime two
Further information: dihedral group:D8
The result does not hold when , a counterexample is the dihedral group of order eight, which is equal to its , even though the exponent is .
Breakdown for higher nilpotence class
Further information: wreath product of groups of order p
For any prime , we can construct a -group generated by elements of order , and with exponent . This is the wreath product of groups of order ; equivalently, it is the -Sylow subgroup of the symmetric group on elements.
It is particularly easy to see from the latter description that the exponent is : there is a cycle of length in the symmetric group, and by Sylow's theorem, some conjugate of this cycle must lie in any Sylow -subgroup. Further, an examination of cycle decompositions shows that there is no element of order .
Further, we can construct a -group generated by elements of order , with exponent for any . This is the iterated wreath product of the cyclic group of order with itself times; equivalently, it is the -Sylow subgroup of the symmetric group on elements. This subgroup contains a cycle of length , and hence, the exponent must be at least .
It suffices to show that the set of elements such that , is a subgroup.
Given: A -group of nilpotence class two, where is an odd prime.
To prove: The set of such that is a subgroup
Proof: Clearly, the identity element is in this set, and if then , so the set is closed under inversion. Thus, what we need to show is that:
Let's prove this. Suppose denotes the commutator of and . Then, since has nilpotence class two, commutes both with and with . Thus, we have:
By induction, we can show that:
Setting , we get:
We also have the identity:
So if , then , and so , and we get: