Elements of the same order in the same conjugacy class that commute are contained in the same cyclic subgroup

Question

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.

Answer 1

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.)

Answer 2

Thanks again for your contribution!

If you are interested, I explain my motivation. In a paper(https://doi.org/10.1080/00029890.2019.1528826) it is proven that a arbitrary group $G$ with a finite number of elements of maximal order has bounded size. Namely: $|G|\leq\frac{mk^2}{\varphi(m)}$, where $m$ is the maximal order and $k$ the number of elements that have order $m$. I wanted to characterize for which groups $G$ the bound is sharp, i.e. $|G|=\frac{mk^2}{\varphi(m)}$. Using GAP I found all groups with this property up to order 1023 and was able to state a conjecture. It is easy to see in the paper, that a group has the property only if all elements of maximal order are conjugated. So we need this as as a requirement.

Conjecture. Let $G$ be a group with $k<\infty$ elements of maximal order $m$, in which all elements of maximal order are conjugated. Then the following are equivalent.
i) $|G|=\frac{mk^2}{\varphi(m)}$
ii) $k=\varphi(m)$
iii) $G$ has a unique subgroup of order $m$
iv) $C_m \cong C_G(x)=C_G(y)\trianglelefteq G$ for all $x,y\in G$ with maximal order

I already proved the equivalence of ii), iii), iv) and ii) $\implies$ i). What I am missing is i) $\implies$ ii). I already proved, that i) implies, that all elements of maximal order commute, but I could not finish till now (this is where my assumption of this post would have helped).