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

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.

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.

• 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!
– 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 $$C_{77}$$.