# 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

### 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.