Omega-1 of odd-order class two p-group has prime exponent

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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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Statement

Let P be a p-group (not necessarily finite) of nilpotence class two, where p is an odd prime. Then, the subgroup:

\Omega_1(P) = \langle x \in P \mid x^p = e \rangle

is a group of prime exponent: in other words, every non-identity element in \Omega_1(P) has order p. Equivalently, in a p-group of nilpotence class two, the product of any two elements of order p is either trivial, or has order p.

Related facts

Generalizations

Facts with similar proofs

Breakdown for the prime two

Further information: dihedral group:D8

The result does not hold when p = 2, a counterexample is the dihedral group of order eight, which is equal to its \Omega_1, even though the exponent is 4.

Breakdown for higher nilpotence class

Further information: wreath product of groups of order p

For any prime p, we can construct a p-group generated by elements of order p, and with exponent p^2. This is the wreath product of groups of order p; equivalently, it is the p-Sylow subgroup of the symmetric group on p^2 elements.

It is particularly easy to see from the latter description that the exponent is p^2: there is a cycle of length p^2 in the symmetric group, and by Sylow's theorem, some conjugate of this cycle must lie in any Sylow p-subgroup. Further, an examination of cycle decompositions shows that there is no element of order p^3.

Further, we can construct a p-group generated by elements of order p, with exponent p^r for any r. This is the iterated wreath product of the cyclic group of order p with itself r times; equivalently, it is the p-Sylow subgroup of the symmetric group on p^r elements. This subgroup contains a cycle of length p^r, and hence, the exponent must be at least p^r.

Proof

It suffices to show that the set of elements x such that x^p = e, is a subgroup.

Given: A p-group P of nilpotence class two, where p is an odd prime.

To prove: The set of x \in P such that x^p = e is a subgroup

Proof: Clearly, the identity element is in this set, and if x^p = e then \left(x^{-1}\right)^p = e, so the set is closed under inversion. Thus, what we need to show is that:

x^p = e, y^p = e \implies (xy)^p = e

Let's prove this. Suppose z = [x,y] denotes the commutator of x and y. Then, since G has nilpotence class two, z commutes both with x and with y. Thus, we have:

xy = (yx)z

And further:

x^2y^2 = xxyy = x(yx)zy = (xy)^2z

By induction, we can show that:

x^jy^j = (xy)^jz^{j(j-1)/2}

Setting j = p, we get:

e = (xy)^pz^{p(p-1)/2}

We also have the identity:

z^p = [x,y]^p = [x,y^p] = [x,e] = e

So if p \ne 2, then p | p(p-1)/2, and so z^{p(p-1)/2} = e, and we get:

e = (xy)^p

as desired.

References

Textbook references