Suppose that $G$ is a group of order $924=2^2\cdot3\cdot7\cdot 11$. Prove that $G$ has an element of order $77$.

My attempt:

By Sylow theorems, we know that there exist elements $ a, b\in G$ with $o(a)=7 $ and $o(b)=11$. Note that $\gcd(7,11)=1$, so if we can show that $ab=ba$ then we are through.

Consider the group $\langle a \rangle$ acting on the set $\Omega=\{g\in G: o(g)=11\}$ by $$ a^k\cdot g:=a^kga^{-k}\ , k=1,2,...,7. $$ Note that the element and its conjugate have the same order and we can easily check that it is a well-defined $\langle a\rangle$ group action on $\Omega$. Now by the Burnside's lemma, we know that the number of orbits, denoted by $|\Omega/\langle a\rangle|$: $$ |\Omega/\langle a\rangle|=\frac{1}{7}\sum_{a^{k}\in\langle a\rangle}|\Omega^{a^k}| $$ where $\Omega^{a^k}=\{g\in\Omega:a^k\cdot g=g\}$.

Now suppose the converse, i.e., there are not elements fixed by $a^k$ in $\Omega$ if $k\ne 7$($a^7=e$ the unit), then $$ |\Omega/\langle a\rangle|=\frac{1}{7}\sum_{e}|\Omega^{e}|=\frac{|\Omega|}{7}\in\mathbb Z. $$ So $\displaystyle 7\vert |\Omega|$. But the number of Sylow $11$-subgroups $n_{11}| 12\cdot 7$ and $n_{11}\equiv 1\pmod {11}$, we have $n_{11}=1$ or $n_{11}=12$ and in either case, $|\Omega|=11-1=10$ and $|\Omega|=12\cdot (11-1)=12\cdot 10=120$, respectively. But neither $7$ divides $10$ nor $7$ divides $120$ and we are done.

Is my reasoning right? Moreover, I am looking for other solutions without using Burnside's lemma. Thank you.


2 Answers 2


Well we know there are elements of order $7$ and $11$ and if any pair of such elements commute then they generate a cyclic subgroup of order $77$.

I think you can argue that if the number of subgroups of order $11$ is not $1$ then it is $12$ (Sylow again: $\equiv 1 \bmod 11$). Take these two cases together.

Take an element $a$ of order $7$ and let it act on these subgroups by conjugation. The orbits must either be single subgroups or sets of $7$ subgroups. In either case there is a subgroup of order $11$ fixed under the conjugation action.

Now consider the action on that subgroup - the automorphism group of a cyclic group of order $11$ has order $10$ and the action induces a homomorphism from the group of order $7$ generated by $a$ to the automorphism group. The image is a subgroup of order $1$ or $7$, and it must be $1$. Therefore $a$ acts trivially on the subgroup of order $11$ and commutes with its members.

  • $\begingroup$ This is awesome! That's exactly where I got stuck when I tried to act $\langle a\rangle$ on the set of groups of order $11$! Thank you so much! $\endgroup$
    – Bach
    May 31, 2019 at 9:57

You can bypass Burnside's lemma completely.

The number of Sylow-11's is $n_{11}=1$ or $n_{11}=12$. Since all Sylow-11s are conjugate, we get $n_{11}=\lvert G\rvert/\lvert N(P_{11})\rvert$ from orbit-stabilizer (as in proof of Sylow first theorem), i.e., $\lvert N(P_{11})\rvert=\lvert G\rvert/ n_{11}$, which is divisible by $7$ in both cases. So there is a Sylow-7 normalizing a Sylow-11, i.e., there is a $C_{11}\rtimes C_7$ as a subgroup of $G$. But there no nontrivial homomorphism $C_7\to\operatorname{Aut}C_{11}$ so this is just $C_{77}$.


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