What is the meaning of group extension?

Short exact sequences

That's what's called a "short exact sequence", and the notation only really makes sense in light of exact sequences more generally. We say a sequence of groups and homomorphisms between them like

$$ \dots \to G_{-1} \to G_{0} \to G_{1} \to G_{2} \to \dots $$

is exact if the kernel of the map from $G_k$ to $G_{k+1}$ is the image of the map from $G_{k-1}$ to $G_k$ for each $k$. This turns out to be a useful concept, but explaining why it's useful is, I think, out of scope.

A special case of this is an exact sequence like the one in your question

$$ 1 \to N \xrightarrow{f} G \xrightarrow {g} Q \to 1.$$

The claim that this sequence is exact tells you three things:

(1) The image of the map $1 \to N$ is the kernel of the map $N \xrightarrow{f} G$. Since the image is trivial that means $f$ is injective.

(2) The image of the map $N \xrightarrow{f} G$ is the kernel of the map $G \xrightarrow {g} Q.$ That is, $\operatorname{im} f$ (which is a copy of $N$ in $G$) inside $G$ is $\ker g$.

(3) The image of the map $G \xrightarrow {g} Q$ is the kernel of the map $Q \to 1$. Since the kernel is all of $Q$, that means $g$ is surjective.

So this can be summarized as follows: The sequence

$$ 1 \to N \xrightarrow{f} G \xrightarrow {g} Q \to 1.$$

is said to be a short exact sequence if $f$ is injective, $g$ is surjective, and $\operatorname{im} f = \ker g$. The only special role that the $1$'s play in this is that they are convenient ways to indicate that $f$ is injective and that $g$ is surjective, given that we are talking in terms of exact sequences anyway. The fact that these claims fit nicely into the framework of exact sequences is part of what makes the framework useful, but it's not useful in an obvious way until you've worked with it a bit, especially if you're only touching short exact sequences.

Special examples of this:

• If $g:G \to H$ is any surjective map, then $1 \to \ker g \xrightarrow{f} G \xrightarrow {g} H \to 1$ is a short exact sequence, where $f: \ker g \to G$ is the inclusion map.

• If $N$ is a normal subgroup of $G$, then $1 \to N \xrightarrow{f} G \xrightarrow {g} G/N \to 1$ is a short exact sequence, where $f$ is the inclusion map and $g$ is the obvious projection map.

But you don't have to use short exact sequences to talk about group extensions.

Group extensions

A group extension of a group $H$ is a group $G$ together with a surjective map $\pi: G \to H$. The extension is said to be central if $\ker \pi \subseteq Z(G)$. It's said to be cyclic if $\ker \pi$ is cyclic, etc.

The basic idea is that elements of $G$ are elements of $H$, but "with some extra information attached". If you started with $G$, you could reconstruct $H$ as a quotient: $H \cong G/\ker \pi.$

The nature of the extra information is, of course, related to what information you forget when you map elements of $G$ down to elements of $H$, i.e. information about the kernel $\ker \pi$. That's why a lot of data about the extension is stated in terms of the kernel: roughly the messier the kernel (and its relationship to $G$), the messier the information you're attaching. The simpler the kernel (and its relationship to $G$), the simpler the information you're attaching.

Group extensions are a basic way of building more complicated groups out of simpler ones. In that sense they're a bit like products of groups, but messier and richer.

The first $1$ indicates that the map $i:N\to G$ is injective. The last that the map from $G$ to $Q$ is surjective. This is because the sequence is exact. That's the kernel of each map is the image of the previous one.

The sequence is called a short exact sequence, and an extension is a super-group, or a group that contains another as a subgroup.

We get $G/i(N)\cong Q$ by the first isomorphism theorem.