In a certain exercise am I asked to show that a group of order $231$ has always an abelian subgroup of order $33$. It might be simple, but I still get struct at the same point.
Here is my approach:
Let $n_p$ be the number of $p$-Sylow subgroups, given primes $p$ that divide the order of the group. In this case, $p$ can be either $3,7$ or $11$. Using Sylow's third theorem I obtained that $n_{11} = 1$ and that $n_3$ can be either $1$ or $7$. If $n_3 = 1$, it is really easy since then both Sylow subgroups for those primes are normal, the intersection is trivial and they form a direct product, and since their orders are prime and coprime between each other, they are cyclic and isomorphic to the cyclic of order $33$, $C_{33}$.
The problem is to show what happens if $n_3 = 7$.
The only thing I have thought about is to show that it is false, using an argument related to counting the elements. However, I'm not really sure how to start, or even if that is true. I would be grateful if someone could give me a hand.
Let $G$ be a group of order $231$. Since $3$ divides the order of the group $G$, it has an element $g\in G$ of order $3$. Now, $\{1,g,g^2\}\subset G$ is an abelian subgroup of order $3$.
I will answer the edited (less trivial) question, that there is a subgroup of order $33$. You already observed that $n_{11}=1$, so there is a normal subgroup $N\subset G$ of order $11$. I constructed above a subgroup $H\subset G$ of order $3$. Now $NH\subset G$ has order $33$. It suffices to check $NH$ is abelian, i.e., that $N$ and $H$ commutes with each other. But $H$ acts trivially on $N$ by conjugation, since the automorphism group of $N$ is $(\mathbb Z/11)^\times\simeq\mathbb Z/10$, which has order coprime to $3$.