# A condition for normalized 2-cocycles from the existence of the inverse element in a group extension

It's a well-known fact that if $$A$$ an Abelian group and $$G$$ is a group, then all group extensions of $$G$$ by $$A$$ is isomorphic with the group ($$A\times G,\,\bullet)$$, where the group operation $$\bullet$$ is

$$(a_1,g_1)\bullet(a_2,g_2) = (a_1+\varphi_{g_1}(a_2)+f(g_1,g_2),\,g_1g_2)\tag{1}$$

where

1. $$\varphi:(A\times G)\to A: (a,g)\mapsto \varphi_g(a)$$ is a group action of $$G$$ on $$A$$
2. $$f: G\times G\to A$$ is a cocycle, i.e. satisfies $$f(g_1,\,g_2g_3)+\varphi_{g_1}(f(g_2,\,g_3)) = f(g_1g_2,\,g_3)+f(g_1,\,g_2)$$.

I'd like to compute the inverse element of $$(a,g)$$. For the sake of simplicity, let's take a normalized $$f$$, that is, for the identity element $$e$$ of $$G$$ suppose $$f(e,e)=0$$. In this case, $$f(g,e)=f(e,g)=f(e,e)=0\tag{2}$$ for all $$g\in G$$ (where $$0$$ is the identity element of $$A$$), and the identity element of $$(A\times G,\bullet)$$ is $$(0,e)$$. So, if $$(a,g)^{-1}=(a_1,g_1)$$ then $$(a_1,g_1)(a,g) = (0,g)\tag{3}$$ and $$(a,g)(a_1,g_1) = (0,g)\tag{4}$$ From (1), (2) and (3) $$g_1=g^{-1}$$ and $$a_1=\varphi_{g}(a)-f(g^{-1},g)\tag{5}$$ while from (1), (2) and (4) $$a_1=\varphi_g(a)+\varphi_g(f(g,g^{-1}))\tag{6}$$

From the equality of the RHS of (5) and (6)

$$-f(g^{-1},g)=\varphi_g(f(g,g^{-1}))\tag{7}$$

Is it sure that this holds for every normalized 2-cocycle? I couldn't derive this from the cocycle condition

$$f(g_1,g_2g_3)+\varphi_{g_1}(f(g_2,g_3)) = f(g_1g_2,g_3) + f(g_1,g_2)\tag{8}$$ Perhaps I missed something?

(I think in (3) and (4), the "$$(0,g)$$" should be "$$(0,e)$$".)
I am not sure how you are getting (5) and (6). Using (1) and (3), I get \begin{align} a_{1}+\varphi_{g^{-1}}(a) + f(g^{-1},g) = 0 \tag{5'} \end{align} and using (1) and (4), I get \begin{align} a+\varphi_{g}(a_{1}) + f(g,g^{-1}) = 0 \tag{6'} \end{align} and applying $$\varphi_{g^{-1}}$$ to (6') gives \begin{align} \varphi_{g^{-1}}(a)+a_{1} + \varphi_{g^{-1}}(f(g,g^{-1})) = 0 \tag{6''} \end{align} and comparing (5') and (6'') gives \begin{align} f(g^{-1},g) = \varphi_{g^{-1}}(f(g,g^{-1})) \tag{7'} \end{align} which you can get by setting $$(g_{1},g_{2},g_{3}) = (g^{-1},g,g^{-1})$$ in (8) and using (2).