Proving a group of order $77$ has a subgroup of order $7$ without Sylow theorem. The question is 

Show if $G$ has order 77 then $G$ has a subgroup of order 7. Without using Sylow Theorems. 

Attempt sketch:
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.
 A: All elements of $G$ must have orders $1$, $7$, $11$, $77$ by Lagrange's Theorem. If $G$ has an element of order $7$, then we're done. If it has an element $a$ of order $77$, then it also has an element of order $7$, namely $a^{11}$.
Therefore we may assume that all elements besides $e$ have order $11$. In that case, $G$ is the union of several subgroups of order $11$, whose pairwise intersections are all $\{e\}$. Thus the cardinality of $G - \{e\}$ must be a multiple of $10$. But this number is actually $76$, a contradiction.
A: Let $a(\neq e)\in G$ then order of an element divides order of the group ,so possible values of $o(a)=1,7,11,77$. 
Now it is not possible that $G$ has all elements of order $11$ because if so then let $o(a)=11,o(b)=11$;take $H=\langle a \rangle  ,K=\langle b \rangle  $ as $o(H)=o(K)=11 $ and $H\cap K=\{e\}$ then $o(HK)=\dfrac{o(H)o(K)}{o(H\cap K)}=121$.Now $HK\subseteq G$ so do you see a contradiction?
Thus $G$ must have atleast one element of order $7$.
If $o(a)=7\implies $ we have a cyclic group generated by $\langle a \rangle  $ of order $7$.
If $o(a)=77$.Then $G$ is cyclic and so $G$ has a unique subgroup corresponding to its each divisor.
DONE.
