I'm trying to understand why $H^2(C_p; \Bbb Z/p)$ despite having $p$ elements there are only two distinct central extensions of $C_p$ by $\Bbb Z/p$. My lecture notes says:
In particular, $H^2(C_p;\Bbb Z/p)$ has $p$ elements. On the other hand in a different sense, there are always two answers, not $p$ of them, namely $C_p \times C_p$ and $C_{p^2}$. The paradox is resolved by considering the specific meaning of when two central extensions are the same. Namely, if you are given
$$1 \to \exp(A) \to G' \to B \to 1$$ $$1 \to \exp(A) \to G \to B \to 1$$
then you have to find an isomorphism between $G$ and $G'$ that is the identity on both $A$ and $B$. In the case that $\exp(A) = B = C_p$, the ability to permute $A$ and $B$ with automorphisms yields two orbits of central extensions. One orbit is a fixed point that gives you $C_p \times C_p$; the other orbit has size $p-1$ and gives you $C_{p^2}$.
Questions:
I'm not sure what an "isomorphism between $A$ and $B$" being an "identity" on both $A$ and $B$ means. Could someone explain this? Also any ideas on how to find that isomorphism?
What does "the ability to permute $A$ and $B$ with automorphism" mean? Moreover, what exactly is the group action that yields two orbits in this context?
Please let me know if you're aware of some reading material or textbook that discusses this topic in detail.
The definition of isomorphism between two extensions of $B$ by $A$ is as follows : a morphism $G\to G'$ such that the following diagram commutes: $\require{AMScd}\begin{CD}1@>>> A @>>> G @>>> B @>>> 1\\ @VVV @VVV @VVV @VVV@VVV\\ 1@>>> A @>>> G' @>>> B @>>> 1\end{CD}$
where the arrow $A\to A $ is $\mathrm{id}_A$ and the arrow $B\to B$ is $\mathrm{id}_B$. So if you see $A$ as a subgroup of both $G,G'$, you want the morphism $G\to G'$ to restrict to the identity on $A$ ("be the identity on $A$"), and if you see $B$ as a quotient of them, you want the morphism $G\to G'$ to induce the identity on $B$ upon quotienting ("be the identity on $B$")
The idea is that if you have an extension $1\to A\to G\to B\to 1$ of $G$, then you can "twist" it by an automorphism of, say, $B$ by the following procedure: compose $G\to B\to B$ (where the second map is the chosen automorphism).
Now this gives an a priori different extension (the map $G\to B$ is no longer the same), and in general, it won't even be isomorphic to the initial extension! (In the diagram above, $B\to B$ has to be the identity, whereas if you did the naive thing here, it would be the automorphism of $B$.)
So given an extension and an automorphism of $B$ (or $A$, both work in a similar way) you get a new extension. Now of course, it may happen that composing with the automorphism does produce an isomorphic extension: if you compose with an automorphism of $A$, it essentially means you can extend that automorphism to $G$ without changing the quotient; if you compose with an automorphism of $B$, it essentially means you can lift that automorphism to an automorphism of $G$ that fixes $A$.
You can see that these are quite strong conditions and won't, in general, be met.
So here the point is that the group $G$ that sites insides the extension has only two possibilities: $C_{p^2}$ or $C_p\times C_p$, but there are various ways it can sit inside the extension, so various extension.
Of course if you have $C_p\times C_p$, then it satisfies the conditions I mentioned, and so twisting the extension by automorphisms of $A$ or $B$ won't change the extension.
But with $C_{p^2}$, the conditions do not hold and so you have an action of $\mathrm{Aut}(B)$ on extensions with $C_{p^2}$ as middle group which is not trivial. In fact, for any $k$ prime to $p$, you can check that $1\to C_p\to C_{p^2}\overset{k}\to C_p\to 1$ is isomorphic as an extension to $1\to C_p\to C_{p^2}\to C_p\to 1$ if and only if $k=1 \bmod p$.
It is interesting to check that here the action of $\mathrm{Aut}(B)$ is the same as the action of $\mathrm{Aut}(A)$.