6
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Where did you find this "fact"? It's not true in general. $\endgroup$
    – Derek Holt
    Dec 22 '20 at 16:53
  • $\begingroup$ 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. $\endgroup$
    – Phil
    Dec 22 '20 at 16:55
  • 1
    $\begingroup$ I believe there is a counterexample of order $75$ with $m=5$. Try $\mathtt{SmallGroup}(75,2)$. $\endgroup$
    – Derek Holt
    Dec 22 '20 at 17:09
  • $\begingroup$ 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. $\endgroup$
    – Phil
    Dec 22 '20 at 17:31
3
$\begingroup$

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.)

$\endgroup$
2
$\begingroup$

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).

$\endgroup$
2
  • 1
    $\begingroup$ 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 $k$. $\endgroup$
    – Derek Holt
    Dec 23 '20 at 13:55
  • $\begingroup$ What I did, thanks for your support. math.stackexchange.com/questions/3959775/… $\endgroup$
    – Phil
    Dec 23 '20 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.