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

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 $$p$$.

Generalizations

 * 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

 * Frattini-in-center odd-order p-group implies p-power map is endomorphism

Breakdown for the prime two
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
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.

Textbook references

 * , Page 183-184, Lemma 3.9, Section 5.3 ($$p'$$-automorphisms of $$p$$-groups)