If $A$ is abelian, then $Ext(G,A)\cong H^2(G,A).$ Like the title says, I want to prove that $Ext(G,A)\cong H^2(G,A),$ where $A$ is an abelian group. This is what I have done so far:
Consider an extension of groups $1\longrightarrow A\longrightarrow M\longrightarrow G\longrightarrow 1$ where $A$ is abelian. Since $M\longrightarrow G$ is surjective it has a set theoretic section $s:G\longrightarrow M.$ Then for each $g_1,g_2\in G,$ there exists $a(g_1,g_2)\in A$ such that $$s(g_1)s(g_2)=a(g_1,g_2)s(g_1g_2).$$ Then it can be proved that $a:G\times G\longrightarrow A$ is a $2-$cocycle. Similarly, if
$t:G\longrightarrow M$ is another choice of section, we get another $2-$cocycle $b:G\times G\longrightarrow A$ such that $$t(g_1)t(g_2)=b(g_1,g_2)t(g_1g_2).$$
Since $s$ and $t$ are both sections of $M\longrightarrow G,$ there is a map $\alpha:G\longrightarrow A$ such that $$t(g)=\alpha(g)s(g).$$
Now, after some calculations it can be proved that $a$ and $b$ have the same class in $H^2(G,A).$ Now we have associated to the extension $M$ of $G$ by $A$ an element of $H^2(G,A).$ I was able to prove that this map is a bijection. Now I only need to prove that this is a group homomorphism, but I am struggling with this. I would appreciate it if someone could help me with this. Thank you for your time!
Note: I am new to group cohomology so I'm only aware of basic things. If you could point me in a direction where I can get a better understanding of these things, I'd really appreciate that.
 A: The group operation of ${\rm Ext}(G,A)$ is given by the Baer sum:
If $1\to A\to M\overset f\to G\to 1$ and $1\to A\to N\overset g\to G\to 1$ are extensions, then first we form the pullback $P$ of $f$ and $g$:
$$P=\{(m,n):f(m)=g(n)\}\subseteq M\times N$$
and then we quotient it out by the normal subgroup $A^*$ generated by $\{(a,a^{-1}):a\in A\}$ so that $[m,an]=[(a,a^{-1})(m,an)]=[am,n]$ in $P/A^*=:U$ for any $a\in A,\,m\in M,\,n\in N$ where $[m,n]$ denotes the $A^*$-coset of $(m,n)$.
Finally, the induced short exact sequence $1\to A\overset\iota\to U\to G\to 1$ will be the sum of the original two extensions.
Now, let $s:G\to M$ and $t:G\to N$ be set theoretic sections with $2$-cocycles $a$ and $b$, respectively.
Then, $[s,t]:=g\mapsto [s(g),\,t(g)]\,\in U$ is a set theoretic section of $U\to G$ which satisfies
$$[s,t](g_1)\cdot [s,t](g_2)\ =\ [s(g_1)s(g_2),\ t(g_1)t(g_2)]\ =\ \\
=\ [a(g_1,g_2)s(g_1g_2),\ b(g_1,g_2)t(g_1g_2)]\ =\ [b(g_1,g_2)a(g_1,g_2)s(g_1g_2),\ t(g_1g_2)]\ = \\
=\ \iota\big(a(g_1,g_2)b(g_1,g_2)\big)\cdot[s,t](g_1g_2)\,.$$
