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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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

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

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: