Normal subgroup $N$, subgroup $U$, then $UN/N = U/N$. Let $G$ be a group and $N \unlhd G$ a normal subgroup, $U \le G$ some subgroup. Then I guess $U / N$ is always some group, and moreover $U / N = UN / N$, because $UN / N = \\{ unN : u \in U, n \in N \\} = \\{ uN : u \in U \\}$ by $unN = uN$ for $n \in N$. Is this right? I am wondering because nowhere I find such an statement that $U / N = UN / N$, moreover, the second isomorphism theorem states something weaker, namely $U / (U\cap N) \cong UN/N$. So I guess such an statement would be interesting, and also writing $U/N$ instead of $UN/N$ seems to be more clearly to me, or have I overlooked something?

As far as I have always understood this, it is completely possible and common for $N$ to not be contained in $U$ at all. So, it would not make sense to mod out by $N$. But, taking $UN$ first, will be the smallest subgroup that contains both $U$ and $N$, which then gives us what we need to make sense of the quotient.

Moreover, the second Isomorphism theorem gives us that $UN/N \cong U/(U\cap N)$.