Does a group with $|G| = 33$ have to contain an element of order $11$? 
A group with $|G| = 33$ must contain an element of order $11$. Prove or disprove.

This is inspired by another MSE question. So we know that there must be an element with order 3.
I tried using Lagrange's theorem and similar, but no luck so far.
EDIT: (context clarification) I would prefer a solution that would not rely directly on Cauchy's theorem - some solution similar to answers to another question I linked to. Let's say this was a problem from a math class, and Cauchy's theorem had not been mentioned.
 A: I'm not sure if this meets your goals. Relying heavily on Lagrange and basic bits about conjugacy classes.
Assume that a group $G$ with $33$ elements, none of order eleven, exists.
If the group had an element $x$ of order $33$, then $x^3$ would be of order $11$, so no such $x$ exists. By Lagrange, all the non-identity element of $G$ must have order three.
Assume first that $G$ has a non-trivial center. Let $a\neq 1$ be a central element. It generates a normal subgroup $N$ of order three. The quotient group $G/N$ is then of order $11$, and thus cyclic. If $xN\in G/N$ is of order $11$, then the order of $x$ in $G$ must be divisible by $11$ - a case we already dealt with.
So we know that $G$ must have a trivial center. Let $a$ be any element of order three. It generates a subgroup $H$ of order three that centralizes $a$. Because $a$ is not in $Z(G)$, we know that the centralizer $C_G(a)=\{x\in G\mid xa=ax\}$ is a proper subgroup of $G$ containing $H$. By Lagrange there cannot be any subgroups properly between $H$ and $G$, so this implies that $C_G(a)=H$.
This implies that $a$ has exactly $11=[G:H]$ conjugates. So all the non-identity elements are partitioned into conjugacy classes of size eleven. But there are 32 of them, so this cannot be. QED

Of course, the class equation is lurking in there, so this comes very close to using Cauchy.
A: Without using Cauchy's theorem one way to approach this problem is with the more powerful Sylow's Theorem. If $\vert G \vert = 33$ then we know that $G$ has at least one Sylow $p$-subgroup for every prime $p$ that divides $G$. Specifically looking at $p = 11$ there is a subgroup $K$ of $G$ with $\vert K \vert = 11$ (the maximal size for a subgroup of order $11^n$ in $G$). Now the only group of order $11$ is $\mathbb{Z}_{11}$ so we know that $K \cong \mathbb{Z}_{11}$. therefore there are elements of $K$ (all the non identity elements) that have order $11$.
A: Sylow's theorems say since $33=3\cdot 11$, then there exists a Sylow $p$-Subgroup of order 11 and that the number of these groups is congruent to $1\pmod {11}$ and divides 33. This means that the group of order 11 is unique. Since this subgroup is of prime order, it is generated by a single element of order 11 in $G$.
