Intuition about the second isomorphism theorem

Question

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:

1. The product $SN$ is a subgroup of $G$.
2. The intersection $S\cap N$ is a normal subgroup of $G$.
3. The quotient groups $SN/N$ and $S/(S\cap N)$ are isomorphic.

I've seen this theorem some time now and I still can't grasp an intuition for it. I mean, it certainly is one important result, because as I've seen 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?

Answer 1

We have a surjective homomorphism $$f : S \to \frac{SN}{N}$$ given by $f(s) = sN$. We have $\ker(f) = S \cap N$, so $$\frac{S}{S \cap N} \cong \frac{SN}{N}$$ In other words, if $f$ is not injective, we quotient out by the kernel to obtain an isomorphism, exactly as we do to prove the first isomorphism theroem. In other words, we would like each coset $sN \in SN/N$ to correspond to $s \in S$. But if $s \in N$, then $sN = N$, so it instead corresponds to a coset $s(S \cap N) \in S/(S \cap N)$

Answer 2

Suppose you drop condition that $N$ is normal in $G$. Then $S,N$ are simply subgroups of $G$. In this case, we can say only about equality of number of cosets. $$|SN\colon N| = |S\colon S\cap N|.$$ But when $N$ is normal, then we can certainly talk about quotient, and it is not only by $N$ but also with some other subgroup, and also isomorphism between them (which are statements (1), (2), (3) in question). I think, this situation can be shown better through diagram:

If $N$ is normal in $G$, then $N$ should be normal in every subgroup in which it is contained. So, if $S$ is other subgroup, then $N$ is certainly contained in $SN$ and hence $N\trianglelefteq SN$ (left part diagram). The isomorphism theorem you concerned says, then $S\cap N$ is then normal in $S$ (right part diagram) and the corresponding quotient groups (think like-red line sections) are isomorphic.

Proving this isomorphism is elementary algebra; no need to think of any strange map; it is most natural one which everyone can think and so it is, in my opinion, the diagram than the proof of this theorem to be understood in the beginning.