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Consider an extension $E$ of a group $G$ by an abelian group $A$. $$1 \to A \overset{\iota}{\to} E \overset{\pi}{\to} G \to 1$$ Two special kinds of extensions are:

  • Central Extensions: $A$ is contained in the centre of $E$.
  • Semidirect products: $\pi$ has a section, i.e. a homomorphism $s : G \to E$ with $\pi \circ s = \mathrm{id}_G$.

It perhaps reasonable to think of these two types of extensions as "orthogonal" since an extension is both central and a semidirect product if and only if it is split, i.e. there is an isomorphism $E \cong A \times G$ through which $ \iota$ and $\pi$ become identified with the standard inclusion and projection.

These two types of extensions are nice in the sense that we can construct all of them in terms of certain external data, namely 2-cocycles and actions.

  • Given $\psi:G \times G \to A$ satisfying $\psi(g_1,g_2) \psi(g_1 g_2,g_3) = \psi(g_1,g_2 g_3)\psi(g_2,g_3),$ we can devise an extension $E_\psi = A\times G$ with product $(a_1,g_1)(a_2,g_2)=(a_1a_2\psi(g_1,g_2),g_1g_2),$ and $\iota,\pi$ given by the standard inclusion and projection.
  • Given a homomorphism $\theta:G \to \mathrm{Aut}(A)$, define $E=A \rtimes_\theta G$ to be $A \times G$ with product $(a_1,g_1)(a_2,g_2) = (a_1 \theta_{g_1}(a_2),g_1g_2)$.

Since the cases of central extensions and semi-direct product are somehow "orthogonal", I am tempted to ask the following ill-defined question:

Main Question: Can we think of all extensions $E$ of a group $G$ by an abelian group $A$ as being somehow built out of these two orthogonal cases of central extensions and semidirect products?

and maybe

Followup Question 1: Can arbitrary extensions be constructed out of external data in the same way as central extensions and semidirect products? The data would need to be something which mixes the notions of a 2-cocycle and an action.

One further question, a bit frivolous, just popped into mind.

Followup question 2: If we want to study extensions $A \to E \to G$ where $A$ is nonabelian, we can still talk about the extensions which are semidirect products. However, the notion of a central extension no long makes any sense. Is there a property $(P)$ which is an appropriate analogue of central extension in this context? Can $(P)$ be chosen so that the trivial split extension is the only extensions which satisfies $(P)$ and is also a semidirect product?

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  • $\begingroup$ I don't really have much of an answer. But for the nonabelian case, note that $2$-cocycles are really elements in the second cohomology group, and when the action is non-trivial,, this generalizes central extensions to abelian extensions. And there is in fact also something called nonabelian cohomology which deals with the more general case, but the theory is much less nice, as one no longer gets cohomology groups, but pointed sets. $\endgroup$ – Tobias Kildetoft Aug 8 '17 at 7:21
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Actually this might be pretty straightforward. Let $1 \to A \overset{\iota}{\to} E \overset{\pi}{\to} G \to 1$ be any extension of $G$ by an abelian group $A$. Since $A$ is normal in $E$, any $e \in E$ determines an automorphism $a \mapsto eae^{-1}$. Moreover, since $A$ is abelian, replacing $e$ by $ea_1$ for some $a_1 \in A$ does not change this automorphism, i.e. the action of $e$ on $A$ only depends on the coset $eA$. Thus, we do in fact have a well-defined action \begin{align*} \theta:G \to \mathrm{Aut}(A) && && \theta_g(a) = eae^{-1} \text{ for any } e\in \pi^{-1}(g), \end{align*} (it is easy to check $g \mapsto \theta_g$ is still a homomorphism)

Next we need something like a 2-cocycle. For every $g \in G$, choose arbitrarily an $e(g)$ in the coset $\pi^{-1}(g)$. Since $e(g_1) e(g_2)$ and $e(g_1g_2)$ are equal under $\pi$, we get a map \begin{align*} \psi : G \times G \to A && \text{ defined by } &&e(g_1)e(g_2) = \psi(g_1,g_2) e(g_1g_2) \end{align*} To see what happens with the 2-cocycle identity, we should compute $e(g_1)e(g_2)e(g_3)$ in two different ways. Because $A$ is not necessarily central in $E$, things come out slightly differently. \begin{align*}e(g_1)e(g_2)e(g_3) &=\psi(g_1,g_2)e(g_1g_2)e(g_3) \\ &= \psi(g_1,g_2)\psi(g_1g_2,g_3) e(g_1g_2g_3) \end{align*} \begin{align*}e(g_1)e(g_2)e(g_3) &=e(g_1)\psi(g_2,g_3)e(g_2g_3) \\ &= e(g_1)\psi(g_2,g_3)e(g_1)^{-1} e(g_1)e(g_2g_3) \\ &= \theta_{g_1}(\psi(g_2,g_3)) \psi(g_1,g_2g_3) e(g_1,g_2g_3) \end{align*} so we find that $\psi$ satisfies the identity $$\psi(g_1,g_2) \psi(g_1g_2,g_3) =\psi(g_1,g_2g_3) \theta_{g_1}(\psi(g_2,g_3))$$ a rather mild variation on the 2-cocycle identity. For convenience, we had may as well require that $e(1)=1$ and that $e(g^{-1}) =e(g)^{-1}$, since this is easy to enforce. This leads to the additional identities $$\psi(g,g^{-1})=\psi(g,1)=\psi(1,g)=1$$

Now we just need to turn all this around:

Let $G$ be a group, $A$ an abelian group. suppose that we have the following data:

  • A homomorphism $\theta : G \to \mathrm{Aut}(A)$
  • a map $\psi : G \times G \to A$ satisfying $\psi(g_1,g_2) \psi(g_1g_2,g_3) =\psi(g_1,g_2g_3) \theta_{g_1}(\psi(g_2,g_3))$ and $\psi(g,g^{-1})=\psi(g,1)=\psi(1,g)=1$ (this second requirement can most likely be dispensed with, at the cost of complicating the formulas).

Define $E$ to be $A \times G$ as a set with product given by $$(a_1,g_1)(a_2,g_2) = (a_1 \theta_{g_1}(a_2) \psi(g_1,g_2),g_1g_2).$$ (This group law is invented by thinking of $(a,g)$ as $ae(g)$ for $g \mapsto e(g)$ some set-theoretic section of $\pi$ defining $\psi$, and multiplying accordingly). There are some routine algebraic checks to show that this defines a group product:

  • $(1,1)(a,g)=(1 \theta_1(a) \psi(1,g),1g)=(a,g)$
  • $(a,g)(1,1) = (a \theta_g(1) \psi(g,1),g1)=(a,g)$
  • $(\theta_g^{-1}(a^{-1}), g^{-1})(a,g)=(\theta_g^{-1}(a^{-1})\theta_g^{-1}(a)\psi(g^{-1},g), g^{-1}g)=(1,1)$
  • $(a,g)(\theta_g^{-1}(a^{-1}), g^{-1})=(a \theta_g \theta_g^{-1}(a^{-1}) \psi(g,g^{-1}), gg^{-1}) =(1,1)$

and, finally, \begin{align*} [(a_1,g_1)(a_2,g_2)](a_3,g_3) &= (a_1 \theta_{g_1}(a_2)\psi(g_1,g_2),g_1g_2)(a_3,g_3) \\ &=(a_1 \theta_{g_1}(a_2) \psi(g_1,g_2)\theta_{g_1g_2} (a_3)\psi(g_1g_2,g_3), g_1g_2g_3) \\ &=(a_1 \theta_{g_1}(a_2)\theta_{g_1} \theta_{g_2}(a_3) \psi(g_1,g_2)\psi(g_1g_2,g_3), g_1g_2g_3) \\ &=(a_1 \theta_{g_1}(a_2)\theta_{g_1} \theta_{g_2}(a_3) \psi(g_1,g_2g_3) \theta_{g_1}(\psi(g_2,g_3)), g_1g_2g_3) \\ &=(a_1\theta_{g_1}(a_2\theta_{g_2}(a_3)\psi(g_2,g_3))\psi(g_1,g_2g_3),g_1g_2g_3) \\ &=(a_1,g_1)(a_2 \theta_{g_2}(a_3)\psi(g_2,g_3),g_2g_3)\\ &=(a_1,g_1)[(a_2,g_2)(a_3,g_3)] \end{align*}

It is quite obvious that $(a,g) \mapsto g$ is a surjective homomorphism whose kernel $\{(a,1) : a \in A\}$ is a isomorphic to $A$.

Strictly speaking, I did not prove that every extension of a group $G$ by an abelian group $A$ comes from the above construction, but this will surely be the case since, as discussed above, every extension $E$ determines such data $\theta$ and $\psi$ once one chooses a section $g \mapsto e(g) : G \to E$. So, one just needs to check the extension constructed out of that data is isomorphic to the original one.

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  • $\begingroup$ You are basically right. The equivalence classes of extensions of this type for a specified action of $G$ on $A$ are in one-one correspondence with the cohomology group $H^2(G,A)$ for this action, and the cohomology group is by definition the quotient $Z^2(G,A)/B^2(G,A)$, where $Z^2(G,A)$ and $B^2(G,A)$ are the groups of $2$-cocycles and $2$-coboundaries. This is all standard material. $\endgroup$ – Derek Holt Aug 8 '17 at 16:09
  • $\begingroup$ @DerekHolt: Really? So the action $\theta : G \to \mathrm{Aut}(A)$ can be somehow eliminated from the 2-cocycle? From my calculations, it looks like one needs to work with some kind of $Z^2(G,A,\theta)$, where the action appears in the cocycle equation. $\endgroup$ – Mike F Aug 8 '17 at 21:19
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    $\begingroup$ You do need the map $\theta$, but that is hidden in the notation. In the expressions $H^2(G,A)$, $Z^2(G,A)$ and $B^2(G,A)$, $A$ denotes a so-called $G$-module, which is precisely an abelian group equipped with an action of $G$ on $A$, which means a homomorphism $\theta:G \to {\rm Aut}(A)$. $\endgroup$ – Derek Holt Aug 8 '17 at 21:34
  • $\begingroup$ @DerekHolt: Ah, OK thanks a lot. $\endgroup$ – Mike F Aug 9 '17 at 0:51

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