Intuition about the second isomorphism theorem

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?

• This isn't "an answer", but I think of it like this: suppose we have a subgroup of $G$, with a normal subgroup $N$. We might want to know "what happens when we quotient out $N$ from some subgroup $H$". The trouble is, $N$ might not be a subgroup of $H$. So we either quotient $N$ out from the smallest subgroup of $G$ containing $H$ and $N$, or we quotient $H$ by the intersection of $H$ and $N$, and both approaches lead us to "the same place". Commented Apr 12, 2016 at 1:49
• the comment of David Wheeler is expanded upon in an answer here: math.stackexchange.com/questions/722632/…
– D.R.
Commented Apr 21, 2022 at 6:03

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 $$\frac{SN}{N} \cong \frac{S}{S \cap N}$$

• Thanks! This explanation made the theorem “click” for me! Commented May 20, 2020 at 17:59
• what does sn^(-)=s^(-) means here??? I mean sn^bar=s^bar Commented Dec 29, 2020 at 9:07
• @RaunitSingh It means the image inside $SN / N$. Commented Feb 16, 2022 at 20:28

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.

• Just to clarify one thing: in case $N$ and $S$ aren't normal, $SN$ is not necessarily a subgroup of $G$ (it is if and only if $SN=NS$). But $SN$ is still the union of cosets of $N$, so it makes sense to write $|SN:N|$. Commented Jul 16, 2016 at 12:50
• yes; second the isomorphism theorem was about isomorphism for some quotients, for which, normality of one subgroups is necessary and sufficient. Commented Jul 17, 2016 at 5:09

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 $s(S \cap N) \in S/(S \cap N)$

There are two additional facts that, in my opinion, make this somewhat more obvious. First,

• Let $\pi$ be the projection map $G \to G/N$.
• Let $\sim$ be the congruence relation defined by $N$; i.e. $x \sim y$ if and only if $xy^{-1} \in N$.

The first key fact is

$$\pi(S) = (SN) / N$$

where $\pi(S)$ means $\{ \pi(s) \mid s \in S \}$. You can think of $SN$ as the subgroup of everything in $G$ that is congruent (by $\sim$) to an element of $S$.

The second isomorphism theorem states that the right hand side is well defined:

• $SN$ is a subgroup of $G$
• $N$ is a normal subgroup of $SN$

The second key fact is that $\sim$ is a congruence relation on $S$, and $S \cap N$ is the congruence class of zero. So you have

$$S /{ \sim} = S / (S \cap N)$$

where the notation on the left means to take the quotient of $S$ by the congruence relation $\sim$; i.e. it's the set of congruence classes, as usual. The second isomorphism theorem states that this is well defined too:

• $S \cap N$ is a normal subgroup of $S$

Finally, the second isomorphism theorem states

$$\pi(S) \cong S / {\sim}$$

With our interpretations of the two sides, we can easily see this as an application of the first isomorphism theorem.