# Isomorphism theorem misunderstanding

One of the isomorphism theorems states $$(HN) \ / \ N \cong H \ / \ (H\cap N)$$.

I am confused about the first part $$(HN) \ / \ N$$, and whether it is equivalent to $$H \ / \ N$$. Every element in $$HN$$ is $$hn$$ for some $$h \in H, n \in N$$. So then every coset in $$(HN) \ / \ N$$ can be expressed as $$\{hnN \text{ for some } h \in H, n \in N\}$$, but this is just the same as $$\{hN \text{ for some } h \in H\}$$, so why isn't the first isomorphism theorem just

$$H \ / \ N \cong H \ / \ (H\cap N)$$

• Not sure this is the first isomorphism theorem. Apr 18 at 20:58
• @Kan't It is in some literature. Apr 18 at 21:01
• As far as I could survey so far, I didn't find any, @azif00. Could you share/mention a such one, please? Apr 20 at 8:09
• Abstract Algebra by Blair and Beachy Apr 21 at 22:23

The notation $$A/B$$ only makes sense (as a group) when $$B$$ is a normal subgroup of $$A$$. In general, we do not require that $$H$$ contains $$N$$ for the theorem holds.
However, the equality $$HN/N = \{hN : h \in H\}$$ is correct.
• I added "(as a group)" because some people study the set of left cosets, denoted by $A/B$, and the notation makes sense as a group when it coincides with right cosets, denoted $B\setminus A$, so when $B\unlhd A$. Apr 18 at 21:11