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

This question has been partly answered in MSE, see here.

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 I am 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.
From here Derek Holt contributed a good point:
All elements of order $$m$$ commute, the elements of order $$m$$ generate an abelian normal subgroup $$N$$ of $$G$$ of exponent $$m$$. We now prove for an arbitrary $$g \in G$$ with order $$m$$, that $$C_G(g) = N$$.
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 $$|N| = mk/\phi(m)$$.

This is where we could not 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 $$y \in Z$$ be 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 $$|G|=mk=\frac{mk^2}{\varphi(m)}$$.

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.

Step 1: Prove that $$m$$ ist not divisible by $$8$$ or a square of an odd prime.
Step 2: We see that $$N=\langle x \rangle \times M$$ with $$M$$ being an elementary abelian $$2$$-group. If $$4\not|\;\; m$$, then $$B=1$$ and we are finished. Else we show the same, via studying how $$C_G(x^2)$$ acts on $$\hat{N}=\langle x^{m/4}\rangle \times M\leq N$$.