I have a relatively simple question. For $p$ odd, does there exist a non-abelian $p$-group such that for every pair of non-commuting $p$-torsion elements $g$,$h$ their difference $g^{-1}h$ is not $p$-torsion?
As a kind of motivation, we ran a search in GAP for 3-groups and on our sample the following trichotomy occurs:
The group is abelian, or
all $3$-torsion elements commute, or
there exist $g,h$ that are $3$-torsion and do not commute, and $g^{-1}h$ is $3$-torsion.
Note: I previously asked it on MathOverflow, but the question was in the wrong form (twice). So I restated it and decided to ask here firstly.
EDIT: Made a mistake again, thank you for pointing that out. The difference is NOT supposed to be $p$-torsion.
EDIT2: By $p$-torsion here I mean "of order $p$".
I may have a proof that such a group does not exist. Still the proof might be wrong, or too long, or unclear - please advise in case.
Suppose, by way of contradiction, that $G$ is such a group. The subgroup of $G$ generated by its elements of order $p$ still satisfies the hypothesis, so we may assume that $G$ is generated by its elements of order $p$.
As noted in the comments, if all elements of $G$ of order $p$ commute pairwise, then the assumption holds trivially. Since we have assumed that $G$ is generated by its elements of order $p$, we may thus assume that $G$ is non-abelian.
Consider the terms $Z_{i}(G)$ of the upper central series of $G$, and the terms $\gamma_{i}(G)$ of the lower central series.
We first show that there is $t \in Z_{2}(G) \setminus Z(G)$ of order $p$.
Since $G$ is non-abelian, and it is generated by its elements of order $p$, the set $$ \{ j : j > 1 \text{, and there is $x \in Z_{j}(G) \setminus Z_{j-1}(G)$ of order $p$}\} $$ is non-empty. Let $i$ be its minimum.
Suppose by way of contradiction that $i > 2$. Let $x \in Z_{i}(G) \setminus Z_{i-1}(G)$ have order $p$. Let $k$ be greatest such that there is $u \in \gamma_{k}(G)$ such that $[x, u] = x^{-1} u^{-1} x u \notin Z(G)$. (This is where we use the fact that $i > 2$.) In particular we have $[x, [x, u]] \in Z(G)$, as $[x, u] \in \gamma_{k+1}(G)$, so that the group $H = \langle x, [x, u] \rangle$ has nilpotence class at most two. Clearly $x^{u} = u^{-1} x u = x [x, u] \in H$. Thus in $H$ we have $$ [x, u]^{p} = (x^{-1} x^{u})^{p} = x^{-p} (x^{p})^{u} [x^{u}, x]^{\binom{p}{2}} = [[x, u], x]^{\binom{p}{2}} = [[x, u], x^{\binom{p}{2}}] = 1, $$ where we have used the facts that $[x^{u}, x] = [[x, u], x] \in Z(G)$, and $p > 2$, so that $p$ divides $\binom{p}{2}$. Thus $[x, u]$ is an element of order $p$ in a $Z_{j}(G)$, for some $i > j > 1$, contradicting the definition of $i$.
So consider $t \in Z_{2}(G) \setminus Z(G)$ of order $p$. As $t \notin Z(G)$, and $G$ is generated by elements of order $p$, there is $x \in G$ of order $p$ which does not commute with $t$. But in the group $K = \langle x, t \rangle$ of class two we now have $$ (x t)^{p} = x^{p} t^{p} [t, x]^{\binom{p}{2}} = [t, x^{\binom{p}{2}}] = 1, $$ once more as $[t, x] \in Z(G)$ and $p > 2$.