# Group extension

Let $F$ and $G$ be groups. An extension of $G$ by $F$ is a solution to the problem of finding a group $E$ that contains a normal subgroup $F'$ isomorphic to $F$ such that the quotient group $E/F'$ is isomorphic to $G$.

More specifically, an extension $\mathcal{E}$ of $G$ by $F$ is a triple $(E,i,p)$ where $E$ is a group, $i$ is an injective group homomorphism of $F$ into $E$, and $p$ is a surjective homomorphism of $E$ onto $G$ such that the kernel of $p$ is the image of $i$. Often, the extension $\mathcal{E}$ is written as the diagram $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$.

An extension is central if $i(F)$ lies in the center of $E$; this is only possible if $F$ is commutative. It is called trivial if $E = F \times G$ (the direct product of $F$ and $G$), $i$ is the canonical mapping of $F$ into $F\times G$, and $p$ is the projection homomorphism onto $G$.

Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ and $\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G$ be two extensions of $G$ by $F$. A morphism of extensions from $\mathcal{E}$ to $\mathcal{E}'$ is a homomorphism $f: E \to E'$ such that $f \circ i = i'$ and $p' \circ f = p$.

A retraction of an extension is a homomorphism $r: E\to F$ such that $r\circ i$ is the identity function on $F$. Similarly, a section of an extension is a homomorphism $s : G \to E$ such that $p \circ s = \text{Id}_G$.

## Equivalence of Extensions

Theorem 1. Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ and $\mathcal{E}' : F \stackrel{i'}{\to} E' \stackrel{p'}{\to} G$ be extensions of $G$ by $F$. Let $u : E \to E'$ be a morphism of extensions. Then $u$ is an isomorphism of $E$ onto $E'$. In other words, every extension morphism is an extension isomorphism.

Proof. Suppose $a,b$ are elements of $E$ such that $u(a)=u(b)$. Then $$p(a) = (p' \circ u)(a) = (p' \circ u)(b) = p(b) ,$$ so $ab^{-1} \in \text{Ker}(p) = \text{Im}(i)$. Let $g \in F$ be an element such that $i(g) = ab^{-1}$. Then $$e_{E'} = u(a)u(b)^{-1} = u(ab^{-1}) = (u \circ i)(g) = i'(g) .$$ It follows that $g$ is the identity of $F$, so $ab^{-1}$ is the identity of $F$ and $a=b$.

Let $a'$ be in $E'$; since $p$ is surjective, there exists $a\in F$ such that $p'(a') = p(a) = p'(u(a))$. Then $$a' u(a)^{-1} \in \text{Ker}(p') = \text{Im}(i') = \text{Im}(u \circ i) \subseteq \text{Im}(u).$$ Thus there exists $b \in F$ such that $u(b) = a'u(a)^{-1}$; then $u(ba) = a'$. Thus $u$ is surjective. $\blacksquare$

Theorem 2. Let $\mathcal{E} : F \stackrel{i}{\to} E \stackrel{p}{\to} G$ be an extension of $G$ by $F$. Then the following are equivalent:

1. $\mathcal{E}$ is the trivial extension;
2. $\mathcal{E}$ admits a retraction;
3. $\mathcal{E}$ admits a section $s$ such that $s(G)$ lies in the centralizer of $i(F)$.

Proof. If $\mathcal{E}$ is the trivial extension, then the projection onto $F$ and the canonical injection of $G$ into $E$ show that conditions 2 and 3 are satisfied. If $\mathcal{E}$ has a retraction $r$, then the mapping $(r,p): E \to F \times G$ is an extension morphism, so $\mathcal{E}$ is isomorphic to the trivial extension. If (3) holds, then the mapping $(f,g) \mapsto i(f)s(g)$ is an extension morphism $F\times G \to \mathcal{E}$, so again $\mathcal{E}$ is isomorphic to the trivial extension of $G$ by $F$. $\blacksquare$

Note that an extension $F \stackrel{i}{\to} E \stackrel{p}{\to} G$ may be nontrivial, but $E$ may still be isomorphic to $F \times G$.