I had a debate with my professor about this today because he assigned the following problem. It is all very simple, yet it seems to me that the results are contradictory.
Let $G$ be a group and let $A$ be a subgroup of $G$. Let $\phi:G\to\bar G$ be a surjective group homomorphism with kernel $N$. Prove that $AN/N\cong\phi(A)$, and prove $A/(A\cap N)\cong\phi(A)$.
His reasoning was this: using the restriction $\phi|_{AN}:AN\to\bar G$, we see that $AN/N\cong\phi(AN)=\phi(A)$, by the First Homomorphism Theorem.
By the same reasoning, $\phi|_{A}:A\to G$ shows that $A/(A\cap N)\cong\phi(A)$.
Thus we have shown $AN/N\cong A/(A\cap N)$.
My problem is this. It seems clear to me that $AN/N$ is equal to $A/N$. The proof below seems valid and shows this. ($e$ is the identity element).
Let $anN$ be in $AN/N$. Then $anN=aN\in A/N$, so $AN/N\subseteq A/N$.
Let $aN$ be in $A/N$. Then $aN=aeN\in AN/N$, so $A/N\subseteq AN/N$.
So we have $AN/N=A/N$.
Something must be wrong here, though. In the first part, we showed that $AN/N\cong A/(A\cap N)$. Consider the case that $A\cap N=\{e\}$ Then we have $AN/N\cong A$, which doesn't make sense.
Furthermore, in general, $A/(A\cap N)\neq A/N$, but as we also showed, $AN/N\cong A/N$ and $AN/N\cong A/(A\cap N)$.
The problem might be just notational. As a set $AN/N=\{a+N:a\in A\}$, but it makes little sense to write $A/N$ since it might not happen that $A\supseteq N$. Thus, we enlarge $A$ to $AN$, and then $AN/N$ makes sense. Simply put, you cannot quotient by something that is not a subobject.