Show that a group of order $33$ cannot only have elements of order $11$ and $1$. 
Show that a group of order $33$ cannot only have elements of order $11$ and $1$.

Attempt:
Assume $G$ only has elements of order $11$ and that there exists $x \in G$ such that $o(x) = 11$. Because $x^{o(G)} = e$ we have $x^{33} = (x^{11})^3 = e$ and so  $x^{11}$ has order $3$. Contradiction.

Every element $g\in G$ of order $11$ generates a subgroup of $G$ of order $11$. Distinct subgroups of order $11$ intersect trivially, so if $n$ is the number of subgroups of $G$ of order $11$ then the number of elements of $G$ is
$$|G|=n\cdot(11-1)+1=1+10n,$$
which cannot equal $33$, so $G$ cannot have only elements of orders $1$ and $11$.
