In group theory we have the second isomorphism theorem which can be stated as follows:
Let $G$ be a group and let $S$ be a subgroup of $G$ and $N$ a normal subgroup of $G$, then:
- The product $SN$ is a subgroup of $G$.
- The intersection $S\cap N$ is a normal subgroup of $G$.
- The quotient groups $SN/N$ and $S/(S\cap N)$ are isomorphic.
Now, I've seem this theorem some time from now and I still couldn't grasp much intuition about it. I mean, it certainly is one important result, because as I've seem it is highlighted as one of the three isomorphism theorems.
The first isomorphism theorem has a much more direct intuition though. We have groups $G$ and $H$ and a homomorphism $f:G\to H$. If this $f$ is not injective we can quotient out what is stopping it from being injective and lift it to $G/\ker f$ as one isomorphism onto its image.
Is there some nice interpretation like that for the second isormorphism theorem? How should we really understand this theorem?
Best Answer
I assume you are having intuitive difficulties with the third statement of the theorem. Let me try and give an intuitive explanation. Every element of $SN$ is of the form $sn$ with $s \in S$ and $n \in N$. Now in $SN/N$ the $n$'s get 'killed' in the sense that in this group $\overline{sn}=\overline{s}$ for $s \in S$ and $n \in N$. However, we are not left with a group that is isomorphic with $S$, because if $s \in N$, that is if $s \in S \cap N$, then $s$ is also the identity in $SN/N$. So, we are left with $S$, but with the remaining part of $N$ completely filtered out, that is $$\frac{SN}{N} \cong \frac{S}{S \cap N}$$