What does a quotient group $G/N$ look like for $C_4$? $C_4 = \{e, a, a^2, a^3\}$
A normal subgroup of $C_4$ is $C_2 = \{e, a^2\}$
So I am wondering what the quotient group $G/N$ looks like in this case.
Ie. where $G = C_4$ and $N = C_2$.
The right (or left) cosets of $N$ are 
$Ne = \{e, a^2\}$ and 
$Na = \{a, a^3\}$
$G/N$ is the group formed by these cosets so I have it as = $\{ \{e, a^2\}, \{a, a^3\} \}$
Is that right...it seems weird having each element being a set of elements..?
 A: That's right. As long as you know what the whole cosets are, you can write them as $[e]$ and $[a]$, where $[\cdot]$ means "the equivalence class of $\cdot$". Also, now that you know $G/N$ has two elements, you should be able to figure out which named group(s) $G/N$ is isomorphic to.
A: Yes, the elements of $G/N$ are cosets, which are themselves subsets of $G$.  But you should think of each individual coset as a single element of the set $G/N$.  But $G/N$ is not just a set -- it's a group.
The operation is multiplication of cosets in the following way:
$$(g_1 N) \cdot (g_2 N) = g_1 g_2 N.$$
In other words, you multiply two cosets as follows.  You take a representative ($g_1, g_2$) of each coset, multiply the two representatives (in the given order, getting $g_1 g_2$) and then the product $(g_1 N) \cdot (g_2 N)$ is the coset containing $g_1 g_2$, namely $g_1 g_2 N$.  However, it turns out that this product is well-defined (you don't get different answers for different choices of representatives from your coset) only if the subgroup $N$ is normal.
In your example, you can make a Cayley table with the rows and columns given by your cosets.  You should find that the multiplication table looks very familiar -- $G/N$ in your example is isomorphic to the only group of order $2$ there is (up to isomorphism).
