Understanding group extensions

Question

This question is about understand what is the intuition behind the following

Definition: An extension of a group $G$ by the group $A$ is given by an exact sequence of group homomorphisms $$1\longrightarrow A\xrightarrow{\phantom{a}\iota\phantom{a}}E\xrightarrow{\phantom{a}\pi\phantom{a}}G\longrightarrow 1$$

Exactness of the sequence means that the kernel of every map in the sequence equals the image of the previous map. Hence the sequence is exact if and only if $\iota$ is injective, $\pi$ is surjective, the image $\operatorname{im}\iota$ is a normal subgroup, and $$\ker \pi=\operatorname{im}\iota \ (\simeq A).$$

This has been extracted from Schottenloher's "Mathematical Introduction to Conformal Field Theory".

Now, I want to gain some intuition behind this. How should we see group extensions intuitively?

Since $\pi$ is a surjective map, we may view it as a projection from $E$ onto $G$. Its kernel is everything that projects onto the identity. This is isomorphic to $A$. So in a sense it is like the whole of $A$ sits "on top of the identity of $G$" inside $E$.

So is the point somehow like at every point $g\in G$ we are attaching a copy of $A$, somewhat akin to what we do with a fiber bundle?

This seems to be in the way to understand this, but only $\pi^{-1}(e)$ which is isomorphic to $A$ right, not all $\pi^{-1}(g)$?

So in summary is this in the right track to understand extensions of groups? If not what is the right intuition about this definition? Why is it called an extension anyway?

Answer 1

Yes, this is the correct intuition. What's more, If $F\longrightarrow E\longrightarrow B$ is a fiber bundle, then on the level of the fundamental group, we get a short exact sequence

$$1 \longrightarrow \pi_1(F) \longrightarrow \pi_1(E) \longrightarrow \pi_1(B) \longrightarrow 1.$$

(If you know about the fundamental group, it might be worth proving this.)

As you pointed out, in the notation of your question, for $g \in G$ not the identity, the preimage $\pi^{-1}(g)$ is not a group. It's a coset. Namely, if $\tilde g$ is an element of $E$ with $\pi(\tilde g) = g$, then we have that $$\pi^{-1}(g) = \tilde gA = \{\tilde ga : a \in A\}.$$ As a set, there is an obvious bijection $A \to \tilde gA$, but since groups are "pointed objects," this isn't a group isomorphism. (Again, if you like the fundamental group, what goes wrong here is roughly that we need to choose a basepoint, and there's a priori no canonical way to change it.)

Nevertheless, I think the fiber bundle analogy is a good one. Unrelatedly, in some parts of group theory, it's more common to hear (in your notation) that $E$ is an extension of $A$ by $G$. My rough understanding is that this correlates with how much you like the functor $\mathrm{Ext}$ in homological algebra. In your convention, I guess we could pretend that the name extension comes from the name for the functor (although surely it's the other way around). In this other convention, you could think of the term extension as meaning "adding elements to $A$ to make $E$."

What's key to both ideas is that group extensions are one of the main ways to build new groups from old ones. The Jordan–Hölder theorem says that any finite group is either simple, meaning it admits no nontrivial strictly smaller quotients, or it can be built up by group extensions from simple groups in a finite number of steps. Thus if one could understand finite simple groups, and understand all possible group extensions, one would know everything about finite group theory.