Why can quotient groups only be defined for subgroups? I know that for the operation on cosets to be well-defined one requires normality. But why is it a requirement (with $G / N$) that $N$ be a subgroup or even a subset of $G$? Surely all that is required is that $N$ is a normal subgroup of a group $S$ (say), of which $G$ is a subgroup of. In which case $G / N \le S/N$.
 A: For $G$ a group, a congruence relation on $G$ is an equivalence relation $\sim$ on $G$ with the property that, whenever $a \sim a'$ and $b \sim b'$, we have $ab \sim a'b'$. In other words, the group operation on $G$ is well-defined on the level of equivalence classes under $\sim$.
Fact: When $N \subset G$ is a normal subgroup, the equivalence relation induced by the partition of $G$ given by cosets of $N$ is a congruence relation. This is precisely what makes it possible to define the quotient group $G/N$.
Conversely,
Theorem: whenever $\sim$ is a congruence relation on $G$, the equivalence class $N := [e]_\sim$ forms a normal subgroup (where $e$ denotes the identity element of $G$), and $\sim$ is precisely the congruence relation induced by the partition of $G$ by $N$-cosets.
Therefore, the only way to take equivalence classes on $G$ in a manner that the group operation is well-defined is via the construction of a quotient group by a normal subgroup.
Proof: That $N$ is a subgroup is obvious, because if $a, b \sim e$, then $ab$ must be equivalent to $e^2 = e$. That $N$ is normal follows almost as easily; if $g \in G$ and $h \in N$, then $e = g e g^{-1} \sim g h g^{-1}$, so $ghg^{-1} \in N$.
It remains to show that $g \sim h$ if and only if $g h^{-1} \in N$. If $g \sim h$, then of course, $h^{-1} \sim h^{-1}$, so $g h^{-1} \sim h h^{-1} = e$, witnessing $gh^{-1} \in N$. Conversely, if $gh^{-1} \in N$, that means $g h^{-1} \sim e$, and multiplication on the right by $h$ gives $g \sim h$.

In answer to the question that Chill2Macht below points out I haven't addressed properly, namely why we require $N$ to be a subset of $G$:
You are absolutely right! If $G$ is a subgroup of $S$ and $N$ is a normal subgroup of $S$, then we can define a group "$G/N$" to be the set of cosets of $N$ by elements of $G$, i.e., $\{gN \mid g \in G\}$. As you can infer by the scare quotes, this is not standard terminology.
Since the $gN$ determine a congruence relation on $G$, it follows from the above theorem that the group "$G/N$" is a quotient of $G$ by a normal subgroup of $G$, but we can be much more precise.
We have the following facts:


*

*The set $$GN = \bigcup_{g \in G} gN = \{gn \mid g \in G, n \in N\}$$ is a subgroup of $S$.

*$N$ is a normal subgroup of $GN$, so that $GN/N$ is defined (standard terminology). Moreover, we have $GN/N = $ "$G/N$", both as a set and as a group.

*Finally, $N \cap G$ is a normal subgroup of $G$, and there is a canonical isomorphism $$G/(N \cap G) \simeq GN/N.$$


This is the content of the Second Isomorphism Theorem, which is stated on Wikipedia with the roles of $G$ and $S$ interchanged.
Note that the last isomorphism comes from applying the first isomorphism theorem to the canonical map $$G \to S/N,$$ $$g \mapsto gN.$$ The image is precisely $GN/N$, and the kernel is precisely $N \cap G$.
