Let $G$ be a group of order 1210 with a subgroup $H$ of order 121. Show that every element of order 11 is in $H$ 
Let $G$ be a group of order 1210 with a subgroup $H$ of order 121. Show that if $a\in G$ has order 11, then $a\in H$

This was a question on a test I just took and even though I spent almost all of my time on it I could not for the life of me figure it out. How can I do this?
Edit: We did not talk about Sylow's theorem in my class so u unfortunately don't know what that means. It's too bad because the answer that uses it is so short and elegant. 
 A: Consider the left coset space $G/H$, define $ K = (\alpha) $ (note that $|K| = 11$, so $K$ is a p-group) and let $ K $ act on $ G/H $ by left multiplication. The fixed point congruence theorem on the actions of nontrivial p-groups states that $ |G/H| \equiv \textrm{Fix}_K (G/H) \pmod{11}$, and since $ |G/H| = 10 $ we conclude that all of $ G/H $ is fixed by the action of $ K $. In particular, $ H $ is fixed, so that we have $\alpha H = H $ and $ \alpha \in H $.
Edit: Here is a more elementary solution. Assume that $ \alpha \notin H $, then $ (\alpha) \cap H = \{e\} $. This implies that $|(\alpha)H| = |(\alpha)||H| = 11^3 $ (why?) However, $(\alpha)H$ is a subset of $ G $, so its number of elements cannot exceed $ 10 \cdot 11^2 $, contradiction. Thus, after all, we have $\alpha \in H $.
A: Note that $H$ is a Sylow $11$-subgroup of $G$. From the Sylow theorem, we know that the number of Sylow's divides $10$ and is $\equiv 1\pmod {11}$, hence $H$ is the only one. Now every element of order $11$ is in a (hence in the) Sylow group ...
