Assuming $H \unlhd G$, construct the quotient group $G/H$ How do I go about doing this? I know that the definition of a quotient group is that it is the set of left cosets (aH) equipped with the following opeation:
$$
(aH)(bH) = (ab)H
$$
Where the identity is $(1_G \cdot H) = H$.
How do I construct the group from here though? Or is this all the question is asking?
 A: You need to make (state) the connection between the hypothesis that $H\unlhd G$ and the conclusion that your $H/G$ is indeed a quotient group.

$$(aH)(bH) = (ab)H, \;\text {where the identity is}\;\; (1_G \cdot H) = H$$

You've got the pieces; you just need to connect them and verify that you do have a quotient group.
That is, simply verify that $(aH)(bH) = abH$ is well-defined, and that the operation defines a "group structure", to show that it is indeed a group.
This is easily established using the premise that $H\unlhd G$. For any old subgroup $H$ of $G$, it does not necessarily hold that $(aH)(bH) = abH,\;\; a, b \in G.\;\;$ It holds iff $H\unlhd G.$
Note: If you haven't learned that left coset multiplication on a subgroup $H$ is well defined if and only if $H$ is a normal subgroup of a group $G$, you should prove that. And if you haven't learned that the left (right) cosets of a normal subgroup $H$ of $G$ form a group $G/H$, then you need to prove that as well.  
A: That's is, you've constructed it! Now prove 
(1) the operation is well defined, meaning 
$$aH=a'H\,\,,\,\,\,bH=b'H\Longrightarrow (ab)H=(a'b')H$$
and 
(2) the operation actually gives you a group structure on the set of (left or right, it doesn't matter) cosets of H.
