Show that a group with $|G| = 33$ contains an element of order $3$.
Using Lagrange's theorem, the possible orders are $1$,$3$,$11$, and $33$ (from the factors of $33$).
If there is a subgroup containing an element with an order of $33$, then we are done.
There is only one identity element with an order of $1$.
I am currently left with elements of orders $3$ or $11$. What is the best way to show that any element of order $11$ is a contradiction?
Best Answer
If $x$ has order $1$, then $x^1 = e$. But that implies $x = e$, so it must be the identity.
You don't need to show that no element has order $11$, but that all elements having order $11$ is a contradiction.
Assume all non-identity elements of $G$ have order $11$. Let $x \in G$, and consider $\langle x \rangle$. It has order $11$, so there's some $y \in G$ that's not in it. Because the intersection of two subgroups is a subgroup, and $11$ is prime, we use Lagrange's theorem to show $\langle x \rangle \cap \langle y \rangle = \{e\}$.
We claim that for $0 \le i,j < 11$, $x^i y^j$ are distinct. If $x^a y^b = x^c y^d$, then $x^{-c} x^a y^b y^{-d} = x^{-c} x^c y^d y^{-d}$, which reduces to $x^{a - c} y^{b - d} = e$. This implies that $y^{b - d} \in \langle x \rangle$, but since the intersection was trivial, $y^{b - d} = e$, and so must $x^{a - c}$. But that means $a \equiv c \pmod{11}$ and $b \equiv d \pmod{11}$, and so the $x^i y^j$ were distinct.
But that's way too many elements! That's $121$ elements, but there were only $33$ in $G$. By contradiction, some element must have order $\ne 11$, and you've already shown the only other choice is $3$.