Show that a group with $|G| = 33$ contains an element of order $3$ 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?
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?
First of all, if there's a non-identity element that is of order one, then you have 2 distinct identities, which must be identical. Contradiction. 

Now to the main point, look at an element $g \in G$  and the cycle $\langle g \rangle$ it generates. That must have order $|\langle g \rangle| =  3, 11,$ or $33$ since we can safely avoid the cycle of order $1$. If we have $3$ then we're done. If it has order $33$ then $G = \langle g \rangle$ which means $g^{11}$ has order $3$. But if $|\langle g \rangle| = 11$ then by Lagrange's Theorem $|G/ \langle g \rangle| = 3$. This is vacuously cyclic so choose one of the generating elements. It has order $3$.