Show that a group with $|G| = 33$ contains an element of order $3$

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 $3$.

Suppose that all the non-identity elements in $G$ have order $11$. Choose an $x_1 \in G, x_1 \ne e$; then $H_1 = \langle x_1 \rangle$ is a subgroup of order $11$.

Now choose $x_2 \in G - \langle x_1 \rangle$; $x_2$ has order $11$, and we again form $H_2 = \langle x_2 \rangle$. Now as a consequence of Lagrange's theorem, $$H_1 \cap H_2 = \{e\}$$

(there is only one proper subgroup of either $H_1, H_2$); thus $H_1 \cup H_2$ account for $11 + 11 - 1 = 21$ elements in the group.

Rinse and repeat to get $H_3$. Argue as before to conclude that this is almost disjoint from both $H_1$ and $H_2$. This accounts for $11 - 1 = 10$ more elements, and a total of $31$.

Question: How do we get the remaining two elements?

According to Lagrange's Theorem, the possible orders for proper subgroups of $\;G\;$ of the form $\;\langle x\rangle\le G\;,\;\;x\in G\;$ are $\;1,3,11\;$ . This number, of course, is also the order of the element $\;x\;$ .

If all the elements of $\;G\;$ had order $\;11\;$ , and since in a cyclic group of order $\;11\;$ there are exactly $\;10\;$ elements of order $\;11\;$ (why?), we'd get at most $\;10+10+10=30\;$ elements (we can't have more elements of order$\;11\;$ since there's one unique element of order $\;1\;$ ...!) .

Thus, there must be at least one element of order $\;3\;$.

Suppose there is no element of order $3$; then every element has order $1$ or $11$. Define an equivalence relation on $G$, calling two elements $a$ and $b$ equivalent if $\langle a\rangle=\langle b\rangle$. The equivalence class of the identity element contains only that one element. The equivalence class of an element of order $11$ contains exactly $10$ elements. Thus the non-identity elements of $G$ are partitioned into disjoint $10$-element sets. Since $32$ is not divisible by $10$, this is impossible.