What is the intuition behind short exact sequences of groups; in particular, what is the intuition behind group extensions?
I'm sorry that the definitions below are a bit haphazard but they're how I learnt about them, chronologically.
In Johnson's "Presentation$\color{red}{s}$ of Groups," page 100, there is the following . . .
Definition 1: A diagram in a category $\mathfrak{C}$, which consists of objects $\{A_n\mid n\in\Bbb Z\}$ and morphisms $$\partial_n: A_n\to A_{n+1}, n\in \Bbb Z,\tag{6}$$ is called a sequence in $\mathfrak{C}$. Such a sequence is called exact if $$\operatorname{Im}\partial_n=\ker \partial_{n+1},\,\text{ for all }n\in \Bbb Z$$ [. . .] A short exact sequence in the category $\mathfrak{C}_{\Bbb R}$ of right $\Bbb R$-modules is an exact sequence of the form $(6)$ with all but three consecutive terms equal to zero. [. . .]
Also, ibid., page 101, is this:
It is fairly obvious that a sequence
$$0\longrightarrow A\stackrel{\theta}{\longrightarrow}B\stackrel{\phi}{\longrightarrow}C\longrightarrow 0$$
is a short exact sequence if and only if the following conditions hold:
$\theta$ is one-to-one,
$\phi$ is onto,
$\theta\phi=0$,
$\ker \phi\le\operatorname{Im}\theta$.
I'm reading Baumslag's "Topics in Combinatorial Group Theory". Section III.2 on semidirect products starts with
Let $$1\longrightarrow A\stackrel{\alpha}{\longrightarrow}E\stackrel{\beta}{\longrightarrow}Q\longrightarrow 1$$ be a short exact sequence of groups. We term $E$ an extension of $A$ by $Q$.
Thoughts:
I'm aware that semidirect products can be seen as short exact sequences but this is not something I understand yet. My view of semidirect products is as if they are defined by a particular presentation and my go-to examples are the dihedral groups.
Please help 🙂
Best Answer
A short exact sequence $1\rightarrow A\rightarrow E\rightarrow Q\rightarrow1$ is really just a fancy way of saying "$E$ has a normal subgroup $A$ where $E/A\cong Q$". [The sequence also gives the isomorphism $\beta: E/A\rightarrow Q$, while $\alpha$ corresponds to the embedding of the abstract group $A$ as a subgroup of $E$.]
Because you care about presentations: if $A$ has presentation $\langle \mathbf{x}\mid\mathbf{r}\rangle$ and $Q$ has presentation $\langle \mathbf{y}\mid\mathbf{s}\rangle$ then the group $E$ given by the above short exact sequence has presentation of the form: $$ \langle \mathbf{x, y}\mid SW_S^{-1} (S\in\mathbf{s}), \mathbf{r}, \mathbf{t}\rangle $$ where $W_{S}\in F(\mathbf{x})$ for all $S\in\mathbf{s}$, and $\mathbf{t}$ consists of words of the form $y^{-\epsilon}xy^{\epsilon}X^{-1}$ with $x\in\mathbf{x}$, $y\in\mathbf{y}$ and $X\in F(\mathbf{x})$. The intuition here is that relators in $\mathbf{t}$ ensure normality of $A$, and so removing all the $x$-terms makes sense. When they are removed you get the presentation $\langle \mathbf{y}\mid\mathbf{s}\rangle$, because of the relators $SW_S^{-1}$. I will leave you to work out where the maps $\alpha$ and $\beta$ fold in to this description.
The above presentation justifies the term extension of $A$ by $Q$: we've started with a presentation for $A$, and then added in the presentation for $Q$ in a specific way to obtain a presentation for $E$.
For a worked example of the above (with some genuinely amazing applications, both in the paper and in subsequent research), look at the paper Rips, E. (1982), Subgroups of small Cancellation Groups. Bull. Lond. Math. Soc. 14: 45-47. doi:10.1112/blms/14.1.45
It is an interesting question when a presentation of the above form does actually define a group extension. This was studied in the paper Pride, S., Harlander, J. & Baik, Y. (1998). The geometry of group extensions. J. Group Theory, 1(4), pp. 395-416. doi:10.1515/jgth.1998.028