What is the maximal order $f(n)$ of a metacyclic (metacyclic group is the extension of a cyclic group by a cyclic group) transitive permutation group of degree $n$? It can be easily proved that $f(p)=p(p-1)$ for prime number $p$, and the following argument (if it is correct) shows that $f(n)$ divides $n^2\phi(n)$:

For a metacyclic transitive permutation group $R$ of degree $n$, we can write $R=\langle a,b\rangle$ with $\langle a\rangle\trianglelefteq R$. Let $M=\langle a\rangle=Z_m$ and $L=\langle b\rangle=Z_l$. Since the point stabilizer of a transitive permutation group is core-free, we have $R_\omega\cap M=1$ for any point $\omega$. This implies that $M$ is semiregular and hence $$ m~\big|~n. $$ Since the center of $R$, denoted by $Z$, is semiregular, we have $|Z|~\big|~n$. Then $C\cap L\leqslant Z$ and $L/(C\cap L)\lesssim Aut(M)=Z_{\phi(m)}$ where $C$ is the centralizer of $M$ in $R$ yields $$ l~\big|~n\phi(m). $$ Therefore, $ml~\big|~n^2\phi(n)$ and thus $|R|~\big|~n^2\phi(n)$.

However, computation results suggest that $f(n)\leqslant n(n-1)$ for all $n$ and $f(n)=n(n-1)$ only if $n$ is prime.


I think I can prove that $f(n) \le n(n-1)$ with equality if and only if $n$ is prime, but it depends on the following result, for which I have not properly worked out the proof.

Let $H$ be the holomorph of the cyclic group $C_n$ of order $n$ - that is the semidirect product of $C_n$ with its automorphism group. Then the maximum order of elements in $H$ is $n$. I have checked by computer that it is true for $n \le 1200$, and I might try and work out the details later.

Anyway, let's assume that is true. If your normal subgroup $M = \langle a \rangle$ is transitive, then $M$ is self-centralizing in $S_n$, so we get $|R| \le n\phi(n)$.

Otherwise, suppose that $M$ has $n/m$ orbits of length $m$ with $m < n$. Then $b$ must act as a cycle of length $n/m$ on the orbits of $M$, so $c := b^{n/m}$ fixes each orbit of $M$. Furthermore, the actions of $c$ on these orbits are all equivalent, and so the the restriction of $c$ to each orbit has the same order. Since this restriction lies in the normalizer in $S_m$ of a cyclic subgroup of order $m$, it lies in the holomorph of $C_m$ and so by the (claimed) result above, this order is at most $m$. So the order of $b$ is at most $n$ and hence $|R| \le mn \le n^2/2$.

Note that, for $n = 2m$ with $m$ odd, there are examples of order $n^2/2$.


The following complements Derek's proof by showing the upper bound for element orders in the holomorph of $C_n$: Set $R=\mathbb Z/n\mathbb Z$. The holomorph of $C_n$ is $R\rtimes R^\times$. Pick $x=(a,b)\in R\rtimes R^\times$. Then $x^k=(a(1+b+\dots+b^{k-1}),b^k)$. Write $s_k=1+b+\dots+b^{k-1}$. Look at the $n$ terms $s_k$ for $k=1,\dots,n$. If it happens that two of them are equal, say $s_u=s_v$ for $1\le u\lt v\le n$, then $0=s_v-s_u=b^us_{v-u}$, and therefore $s_{v-u}=0$. If however these terms are pairwise distinct, then one of them must be $0$. So at any rate, there is $1\le k\le n$ with $s_k=0$. Furthermore, $0=(b-1)s_k=b^k-1$, so $x^k=1$.


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.