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

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?

• Where did you find this "fact"? It's not true in general. Dec 22 '20 at 16:53
• Maybe "fact" is the wrong word, sorry, I just tried many examples and it always worked. Could you provide me a counterexample? I'm doing some little research and found out using GAP, that this is true for all groups of order $<1024$(no data for order $512;708$ though), if all elements of maximal order are in the same conjugacy class.
– Phil
Dec 22 '20 at 16:55
• I believe there is a counterexample of order $75$ with $m=5$. Try $\mathtt{SmallGroup}(75,2)$. Dec 22 '20 at 17:09
• Thanks Derek, this is indeed a counterexample. Well then I'm interested why this seems to be true, if all elements of maximal order are in the same conjugacy class.
– Phil
Dec 22 '20 at 17:31

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.)

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).

• If you want help with this, then you should ask it as a new question. After thinking about it, I believe your conjecture. In a group satisfying (i), the elements of order $m$ must generate an abelian normal subgroup of index $k$. Dec 23 '20 at 13:55
• What I did, thanks for your support. math.stackexchange.com/questions/3959775/…
– Phil
Dec 23 '20 at 18:26