I have a conjecture, that I would like to prove.
EDIT: My first idea was not true, as pointed out by Derek Holt.
Statement:
Let $G$ be a (finite) group and $m$ the maximal order of an element of $G$, such that all elements of maximal order are conjugated.
If two commuting elements $x,y \in G$ have order $m$, then $x^n=y$ for some $n \in \mathbb{N}$, i.e. $y$ is contained in the cyclic subgroup generated by $x$
I looked for some examples:
$(1,2,3)=(1,3,2)^2\in S_3$
$(1,2,3,4)=(1,4,3,2)^3 \in S_4$, but for non-commuting elements of this conjugacy-class it is not true, see: $(1,2,3,4)\not=(1,3,2,4)^n \forall n \in \mathbb{N}$.
Is this true in general? Does anyone know a proof?
Thanks in advance.
Best Answer
There is a counterexample to your revised question, which is the Frobenius group with structure $11^2:{\rm SL}_2(5)$. The highest order of an element is $11$, and all elements of order $11$ are conjugate. You can access this group in GAP as $\mathtt{PrimitiveGroup}(121,56)$.
I would guess that there are not too many couterexamples, and I wonder whether this might be essentially the only one. (By ``essentially'' I mean that you can get other counterexamples by taking direct products with other groups.)