 # gr.group idea – Abstract proof that \$lvert H^2(G,A)rvert\$ counts group extensions Answer

Hello expensive customer to our community We will proffer you an answer to this query gr.group idea – Abstract proof that \$lvert H^2(G,A)rvert\$ counts group extensions ,and the respond will breathe typical by way of documented data sources, We welcome you and proffer you fresh questions and solutions, Many customer are questioning in regards to the respond to this query.

gr.group idea – Abstract proof that \$lvert H^2(G,A)rvert\$ counts group extensions

(This query is initially from Math.SE, the place it did not obtain any solutions.)
$$CommandMathOperator{Hom}{Hom} CommandMathOperator{im}{im} CommandMathOperator{id}{id} CommandMathOperator{ext}{Ext} newcommand{Z}{mathbb{Z}}$$

Let $$G$$ breathe a bunch, let $$A$$ breathe a $$G$$-module, and let $$P_3to P_2to P_1to P_0toZto0$$ breathe the initiate of a projective decision of the $$G$$-module $$mathbb{Z}$$. Consider the cohomology group

$$H^2(G,A)=frac{ker(Hom_{Z G}(P_2,A)toHom_{Z G}(P_3,A))}{im(Hom_{Z G}(P_1,A)toHom_{Z G}(P_2,A))}.$$

It can breathe proven that $$lvert H^2(G,A)rvert$$ counts the variety of equivalence courses of group extensions $$0to Ato Eto Gto0$$.
The solely proof that I do know of this outcome includes selecting a particular projective decision (particularly, the barrier decision).

Is there a proof of this outcome that doesn’t require selecting a particular projective decision?

For context, $$lvertext_R^n(M,N)rvert$$ counts the variety of equivalence courses of extensions $$0to Nto X_ntoldotsto X_1to Mto0$$. The proof of this outcome is pretty summary and doesn’t require selecting a particular projective decision of $$M$$ or a particular injective decision of $$N$$.

Also, I’m cognizant that we even have isomorphisms in each of those outcomes however I’m extra within the actuality of an categorical bijection.

Here is one method for setting up an component of $$H^2(G,A)$$ from an extension $$0to Ato Eto Gto0$$: Treat $$A$$ as an $$E$$-module and deem the transgression map $$H^1(A,A)^{E/A}to H^2(E/A,A^A)$$.
Rewriting this offers a homomorphism $$Hom(A,A)^Gto H^2(G,A)$$.
The picture of $$id_A$$ underneath this map will breathe an component of $$H^2(G,A)$$.

To make this labor, this map would necessity to breathe a bijection from equivalence courses of group extensions and components of $$H^2(G,A)$$.

Another method that I thought of was to labor immediately with the capricious projective decision (just like the proof of the Yoneda Ext outcome).
Suppose that we’re given a bunch extension $$0to Ato Eto Gto0$$.
We need to assemble an component of $$ker(Hom_{Z G}(P_2,A)toHom_{Z G}(P_3,A))$$.
Equivalently, we need to assemble a $$Z G$$-module homomorphism $$P_2/im(P_3to P_2)to A$$.
However, $$im(P_3to P_2)=ker(P_2to P_1)$$ and $$P_2/ker(P_2to P_1)congim(P_2to P_1)=ker(P_1to P_0)$$.
Thus, we need to assemble a $$Z G$$-module homomorphism $$fcolonker(P_1to P_0)to A$$.
Furthermore, if we unwind some extra definitions, we behold that we solely necessity to assemble $$f$$ as much as the restriction of a $$Z G$$-module homomorphism $$P_1to A$$.

Unfortunately, the one data we now have about $$A$$ is the brief require sequence $$0to Ato Eto Gto0$$ which makes it difficult to outline a $$Z G$$-module homomorphism to $$A$$.

we’ll proffer you the answer to gr.group idea – Abstract proof that \$lvert H^2(G,A)rvert\$ counts group extensions query through our community which brings all of the solutions from a number of dependable sources.