Some of the work I have been doing lately is heavily dependent on chasing commutative diagrams so I have been brushing up on short exact sequences since I was not familiar with them. For the most part I feel I understand them but I have a couple things I want to clarify. In my examples the algebraic structures in my exact sequence's will be groups. In fact I think my misunderstandings might stem from the fact that I only have elementary knowledge of group theory.
Considering the short exact sequence
$$0 \rightarrow A \xrightarrow{\phi} B \xrightarrow{\psi} C \rightarrow 0$$
according to Wolfram
the group $A$ can be considered as a (normal) subgroup of $B$, and $C$ is
isomorphic to $B/A$.
Most sites I checked made analogous statements. But what I in fact derived was only that $C \simeq B / \mathrm{Im(\phi)}$, but also that $\mathrm{Ker}(\psi) = \mathrm{Im(\phi)} \simeq A.$ So I feel I am very close to seeing what they are saying but something is holding me back. Since $A$ is isomorphic to Im($\Phi$) = Ker($\psi$) I see why they would say "$A$ can be thought of as a normal subgroup of $B$" since kernals are normal, specifically the normal subgroup such that $C \simeq B/A$. But I guess what I want to know is, what good does it do to $think$ about it being this way, even though it might not $actually$ be true that $C \simeq B/A$.
I have seen several examples where it is exactly true such as
$$0 \rightarrow 2\mathbb{Z} \hookrightarrow \mathbb{Z} \twoheadrightarrow \mathbb{Z} \, / \, 2\mathbb{Z} \rightarrow 0.$$
But for some reason that seems a bit contrived to me. Maybe because $C$ is exactly the quotient of $B$ and $A$ or because $A$ is exactly a normal subgroup of $B$.
I feel like one could easily find an example where it is not the case, although the general properties I mentioned I derived would still hold. What would this mean? Should you then just actually find the normal subgroup of $B$ that is isomorphic to $A$ and construct a new short exact sequence using this new group in place of $A$?
But there must really be something to this perception, even wiki says
It is helpful to think of A as a subobject of B giving rise to the isomorphism….
Why? Why is that helpful wiki?
Also, I add that I have already seen this similar question What is a short exact sequence telling me? but still felt my question had not been answered.
Best Answer
I understand your confusion. Let me try and explain. Let $$ 0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$$ be a short exact sequence of groups. Recall that exactness in this case is equivalent with saying $f$ is injective, the kernel of $g$ is the image of $f$ and $g$ is surjective. So $A \cong \text{Im}f \subset B$. So $A$ is isomorphic to a subgroup of $B$. So $B$ contains $A$ as a subgroup, up to isomorphism. In fact, we could replace every element of $A$ by its corresponding element of $\text{Im} f$, without changing anything structural. Intuitively I think of it like this: if two groups are isomorphic it means that they are the same 'up to symbols', that is, the one is another way of writing the other, but the group structure (the structure that is independent of how you write the group down) is the same. In this example, $B$ contains a subgroup that is 'essentially' $A$.
With the first isomorphism theorem we also have $B/\text{Im}f \cong C$. In fact, the short exact sequence above is equivalent to $$0 \rightarrow \text{Im} f \xrightarrow{\iota} B \xrightarrow{\varphi} B/\text{Im} \rightarrow 0$$ where $\iota$ is the injection map (it sends $a \in \text{Im} f$ to $a \in B$), and $\varphi$ is the canonical quotient map (it sends $b$ to $\overline{b}$). By the way, two short exact sequences are called equivalent if their corresponding groups are isomorphic and the diagram induced by these isomorphisms commutes. I hope this makes things a bit more clear.