The question is
Show if $G$ has order 77 then $G$ has a subgroup of order 7. Without using Sylow Theorems.
Let $x \in G$. By Lagrange's theorem the order of $x$ is either $1, 7, 11$. Suppose $x \neq e$ Then $x$ has order $7$ or $11$. Now suppose $|x| = 7$. Then $x$ is the generator of a group of order $7$ and we are done. Suppose now that there does not exist an element of order $7$ in G. Then $G$ cannot have order $77$ since $G$ would be cyclic of order $11$ Hence $G$ has an element of order $7$ and thus a subgroup of order $7$.
Please excuse the horrible way this is written.