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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.

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    $\begingroup$ Nice question (following another you posted here). If I would be you I would post this on mathoverflow.net too. $\endgroup$ Dec 23 '20 at 18:55
  • $\begingroup$ Thanks Nicky, I guess I follow your advice after some time here :) $\endgroup$
    – Phil
    Dec 23 '20 at 18:59
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    $\begingroup$ Yes, I think this problem would be suitable for mathoverflow. $\endgroup$
    – Derek Holt
    Dec 25 '20 at 8:42
  • $\begingroup$ OK, I will move on there. I really appreciate all your help! $\endgroup$
    – Phil
    Dec 25 '20 at 9:39
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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 $|N|=p$

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  • $\begingroup$ I think this follows from Lemma $3$ in the cited paper. $\endgroup$
    – the_fox
    Dec 24 '20 at 20:52
  • $\begingroup$ @the_fox Sorry, what follows from Lemma 3 of the paper - I am afraid I have not looked at it! $\endgroup$
    – Derek Holt
    Dec 24 '20 at 21:20
  • $\begingroup$ That if $G$ is abelian and $|G|=mk/\phi(m)$ then $G$ is cyclic and $k=\phi(m)$. $\endgroup$
    – the_fox
    Dec 24 '20 at 21:26
  • $\begingroup$ Ah OK - well I guess that proves that (i) implies (ii) and answers the question! $\endgroup$
    – Derek Holt
    Dec 24 '20 at 22:29
  • $\begingroup$ 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. $\endgroup$
    – the_fox
    Dec 24 '20 at 23:22

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