Bound for order of a group depending on number of elements of maximal order

In a paper On the Number of Elements of maximal order in a Group it is proven that an 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 all groups $$G$$, where the limit 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 this property only if all elements of maximal order are conjugated. So we need this as as a requirement.

I wanted to prove the following conjecture, but missing some tiny part. Maybe someone knows a way, I would be really happy.

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

Proof.
$$i) \implies ii)$$ This is the part, I could not prove:
I only could prove, that all elements of order $$m$$ commute:
Let $$C_G(x)$$ be the stabilizer of an element of maximal order. Orbit-Stabilizer-Theorem tells us, that $$|C_G(x)|=\frac{mk}{\varphi(m)}$$. Assume there exists an element of order $$m$$, not contained in $$C_G(x)$$. $$\langle x \rangle$$ operates via left-multiplication on $$C_G(x)$$. $$C_G(x)$$ is partitioned into $$\frac{|C_G(x)|}{m}$$ orbits. According to Lemma 3 of the paper linked above, in each orbit exist at least $$\varphi(m)$$ elements of order $$m$$, i.e. in $$C_G(x)$$ exist at least $$\varphi(m)\frac{|C_G(x)|}{m}$$ elements of order $$m$$. Our assumption tells us $$\varphi(m)\frac{|C_G(x)|}{m} < k$$, which leads to the contradiction $$|C_G(x)| < \frac{mk}{\varphi(m)}$$. It follows that all elements of order $$m$$ commute.
This is where I can't proceed further. Maybe someone has an idea?

$$ii) \iff iii)$$ If $$k=\varphi(m)$$, an element of order $$m$$ generates a cyclic subgroup which contains $$\varphi(m)$$ elements of order $$m$$, that all generate this subgroup. So there can't be other elements of order $$m$$ in different subgroups. Otherwise, if there is only one cyclic subgroup of order $$m$$, then it contains $$\varphi(m)$$ elements of order $$m$$, no additional elements of order $$m$$ can exist, as they would generate a second cyclic subgroup of order $$m$$.

$$iii) \implies iv)$$ Let $$Z$$ be the unique subgroup of order $$m$$ and $$X=\{x_1,\dots,x_k\}$$ the set of elements of order $$m$$. As all $$x\in X$$ generate $$Z$$, $$Z$$ must be contained in all centralizers of elements in $$X$$. Note that $$G$$ operates on itself via conjugation. Orbit-Stabilizer-Theorem tells us for $$x \in X$$: $$|G|=|^Gx||G_x|=k|G_x|=\frac{mk^2}{\varphi(m)}=mk$$ This follows as all elements of order $$m$$ are conjugated and $$k=\varphi(m)$$ holds. It follows, that $$|G_x|=m$$, which leads to $$G_x=Z\cong C_m$$ for all $$x \in X$$.
For the normal subgroup part, note that $$\phi(x_i)=x_j$$ for an inner automorphism $$\phi$$ and $$i,j\in \{1,\dots k\}$$. Let $$y \in Z$$ arbitrary, then $$y=x_1^\alpha$$ for $$\alpha \in \mathbb{N}$$. Let $$\phi$$ be an arbitrary inner automorphism. It follows that there is a $$i \in \{1,\dots k\}$$ with $$\phi(y)=\phi(x_1^\alpha)=\phi(x_1)^\alpha=x_i^\alpha \in Z$$ It follows that $$Z$$ is invariant under inner automorphisms, i.e. normal.

$$iv) \implies i)$$ Orbit-Stabilizer-Theorem tells us that $$|G|=|^Gx||G_x|=mk$$. As all stabilizers of elements of order $$m$$ are equal to the same cyclic group of order $$m$$, it follows, that there exist only one cyclic group of order $$m$$, it follows $$k=\varphi(m)$$ and $$|G|=mk=\frac{mk^2}{\varphi(m)}$$.

Thanks to anyone, who read till here ;)

Another property, which my GAP-study suggests to be equivalent is :
$$v)$$ $$G'$$ is cyclic
This proof has low priority, as I first want to have my circle-implications. I guess I can show, that $$G'$$ is contained in the unique cyclic group $$Z$$ of order $$m$$, by proving, that $$G/Z$$ is abelian. I did not succeed yet, though.

• Nice question (following another you posted here). If I would be you I would post this on mathoverflow.net too. Dec 23 '20 at 18:55
– Phil
Dec 23 '20 at 18:59
• Yes, I think this problem would be suitable for mathoverflow. Dec 25 '20 at 8:42
• OK, I will move on there. I really appreciate all your help!
– Phil
Dec 25 '20 at 9:39

This is not a complete answer, but provides a bit more information about a group $$G$$ with $$|G| = mk^2/\phi(m)$$.

Since you have proved that all elements of order $$m$$ commute, the elements of order $$m$$ generate an abelian normal subgroup $$N$$ of $$G$$ of exponent $$m$$.

Let $$g \in G$$ have order $$m$$. We claim that $$C_G(g) = N$$. To prove this, let $$h \in C_G(g)$$. We want to show that $$h \in N$$. This is clear if $$h \in \langle g \rangle$$. Otherwise, since $$m$$ is the largest order of any element in $$G$$, $$\langle g,h \rangle$$ is a $$2$$-generator abelian group of exponent $$m$$, and so it is equal to $$\langle g \rangle \times \langle hg^i \rangle$$ for some $$i$$ with $$0 \le i < m$$. But then $$hg^{i+1}$$ has order $$m$$ and hence lies in $$N$$, so $$h \in N$$, which establishes the claim.

So $$[G:N] = [G:C_G(g)] = k$$, and hence $$|N| = mk/\phi(m)$$.

I think that an abelian group $$N$$ of exponent $$m$$ with $$k$$ elements of order $$m$$ can only have order $$mk/\phi(m)$$ when $$k=\phi(m)$$ and $$N$$ is cyclic, and it should be possible to prove this, but I have done so yet. You can easily check for example that if $$m=p$$ is prime then it is only possible when $$|N|=p$$

• I think this follows from Lemma $3$ in the cited paper. Dec 24 '20 at 20:52
• @the_fox Sorry, what follows from Lemma 3 of the paper - I am afraid I have not looked at it! Dec 24 '20 at 21:20
• That if $G$ is abelian and $|G|=mk/\phi(m)$ then $G$ is cyclic and $k=\phi(m)$. Dec 24 '20 at 21:26
• Ah OK - well I guess that proves that (i) implies (ii) and answers the question! Dec 24 '20 at 22:29
• I think I was wrong; spoke too hastily! The claim that if $G$ is abelian of exponent $m$ with $k$ elements of order $m$ and $|G|=mk/\phi(m)$ then $G$ is cyclic and $k=\phi(m)$ does not hold for e.g. $G=C_4 \times C_2$. Am I being silly? (My brain is a little fried from thinking about something else at the moment.) What might be true is that if $G$ is abelian then $k \geq \phi(|G|)$ and equality occurs if and only if there is only one factor of order $m$ in the primary decomposition of $G$. And I am not even sure now that that follows from the lemma I mentioned. Dec 24 '20 at 23:22