Extension of a group $G$ by a commutative group $F$

Question

Let $F,G$ be two groups. An extension of $G$ by $F$ is a triple $\mathscr{E}=(E,i,p)$, where $E$ is a group, $i:F\rightarrow E$ is an injective homomorphism, and $p:E\rightarrow G$ is a surjective homomorphism such that $Im(i)=Ker(p)$.

Denote by $\mathscr{E}:F\xrightarrow{i} E\xrightarrow{p} G$ the extension $\mathscr{E}=(E,i,p)$ of $G$ by $F$.

Let $\mathscr{E}:F\xrightarrow{i} E\xrightarrow{p} G$ be an extension of G by a commutative group $F$. The group $E$ operates on $F$ by inner automorphisms, this operation is trivial on the image of $F$ and hence defines an operation of $G$ on $F$.

What is going on here? Specifically, what is the homomorphism corresponding to the action of $E$ on $F$ by inner automorphisms?

Thank you in advance.

Answer 1

I've just answered a similar question here, but I will phrase it again somewhat more quickly. If you want any more details, please comment and I'll be happy to explain further.

Again, I will write the extension as

$$0 \to K \xrightarrow{i} G \xrightarrow{p} Q \to 0$$

because I think using $K$ for the kernel and $Q$ for quotient helps keep terminology straight.

In the linked question, we required a splitting $s : Q \to G$ in order to define our action of $Q$ on $K$. In this question, we are able to remove that assumption. The price we pay is that we must assume $K$ is abelian. Why is that true? Let's see:

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Here is the snappy proof given by Bourbaki:

Define $\gamma_g(k) : K \to K$ by $\gamma_g(k) = gkg^{-1}$ as before. Now, since $K$ is abelian, notice

$$\gamma_k(k') = k k' k^{-1} = k'$$

This tells us that $\gamma_{-} : G \to \text{Aut}(K)$ (the function taking $g$ to $\gamma_g$) has $\gamma_k = \text{Id}$ for each $k \in K$, and so $\gamma_{-}$ descends to the quotient $G/K$ by the universal property of the quotient. Since $G/K \cong Q$, the claim follows.

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Let's unpack that a little bit. We know $G/K \cong Q$, so we know each $q \in Q$ is isomorphic to $g_q K \in G/K$ for some $g_q \in G$. Now we define our action by

$$\tau_q(k) = g_q k g_q^{-1}$$

Again, if you want to be extra crystal clear, we should actually define it as

$$\tau'_q(k) = i^{-1} \left ( g_q i(k) g_q^{-1}\right)$$

but nobody has time for that.

All we have to do now is check that $\tau_q$ is well defined. What could go wrong if we picked $g_q'$ as a coset representative instead of $g_q$?

Well, $g_q' g_q^{-1} \in K$, so $g_q' = k g_q$. Now we see:

$$g_q' k' g_q^{-1} = (k g_q) k' (k g_q)^{-1} = k (g_q k' g_q^{-1}) k^{-1} = g_q k' g_q^{-1}$$

where the last equality comes from $K$ being abelian (and $g_q k' g_q^{-1} \in K$).

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As an aside, you can drop the hypothesis that $K$ is abelian AND the hypothesis of a splitting $s : Q \to G$ if you are willing to put in some extra work. I won't go into it here, but it is a very interesting (and useful!) topic.

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I hope this helps ^_^

Answer 2

$i$ is injective and the image $i(F)$ is a normal subgroup. Any element $e\in E$ stabilizes this subgroup, so if $f\in F$ then there is an $f'$ such that $ei(f)e^{-1}=i(f')$. This means that the action of $e$ on $f$ yields $f'$. These aren't inner automorphisms of $F$, but the inner automorphisms of $E$ act on the image of $F$, which pulls back to $F$.