Induction based on the 2-cocycle identity for $c': \Bbb Z/p \times \Bbb Z/p \to \Bbb Z/p$ and $c'(0, 0) = 0$

Question

Context:

I was trying to find the number of elements in $H^2(C_p; \Bbb Z/p)$ and my professor gave this hint (cf. Understanding the relation between central extensions and 2-cocycles and coboundaries):

Probably the main trick for a calculation like this is to use coboundaries to clean up a cocycle. Let $b$ be a generator of $C_p$, and let $c$ be any 2-cocycle. Then $c$ is equivalent to another cocycle $c'$ such that $c'(b^x,b^y) = 0$ for all non-negative $x$ and $y$ with $x+y < p$. You can prove this by induction on the value of $x+y$. You cannot make it happen for $x+y = p$. However, after killing $c'(b^x,b^y)$ for $x+y < p$, the cocycle condition then implies that $c'(b^x,b^y)$ is equal for all values of $x+y \geq p$, provided that $0 \leq x,y < p$. That is then the extra dimension in $Z^2(C_p;\Bbb Z/p)$ that is not present in $B^2(C_p;\Bbb Z/p)$.

Question:

Consider the 2-cocycle identity: $c'(y,z)+c'(x,y+z) = c'(x+y,z)+c'(x,y)$ where $c': \Bbb Z/p \times \Bbb Z/p \to \Bbb Z/p$ and $c'(0, 0) = 0$. $p$ is a prime.

I was given an exercise to show that $c'(x, y) = 0$ for all $x+y < p$ and $c'(x, y)$ is equal for all values $x+y \geq p$ provided that $0 \leq x, y < p$. $x$ and $y$ are non-negative integers.

Is there any simple way to show this?

My attempt:

I could prove so far that $c'(1, 0) = 0$ and $c'(0, 1) = 0$ as follows:

If $x= 1, y = 0$ and $x + y = 1$ (let $z=0$):

$$c'(x+y,z) = c'(y,z)+c'(x,y+z) - c'(x,y) \\ \implies c'(1,0) = c'(0,0)+c'(x,0) - c'(x,0) = c'(0,0) = 0 $$

If $x = 0, y=1$ and $x + y = 1$ (let $z=p-1$):

$$c'(x,y) = c'(x,y+z) + c'(y,z) - c'(x+y,z)\\ c'(0,1) = c'(0,1+p-1) + c'(1,p-1) - c'(1,p-1) = c'(0, 0) = 0$$

How should I proceed after this?

Answer 1

In what follows I will often not make a distinction between an integer $x$ and its class in $\mathbb{Z}/p\mathbb{Z}$. I will also write the law in $C_p$ additively, so if $x\in \mathbb{Z}$, I will allow myself to write $x\in C_p$, instead of $b^x$ where $b$ is a generator.

For any $x,y\in \mathbb{Z}$ with $0\leqslant x,y<p$, let us write $\chi(x,y)=0$ if $x+y<p$, and $\chi(x,y)=1$ if $x+y\geqslant p$. This defines a function $\chi: C_p^2\to \mathbb{Z}/p\mathbb{Z}$, and you can easily check that it is a $2$-cocycle. You can either verify directly the cocycle condition, or you can notice that it is the cocycle corresponding to the (non-split) extension $0\to \mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}\to \mathbb{Z}/p\mathbb{Z}\to 0$, with the obvious section $\mathbb{Z}/p\mathbb{Z}\to \mathbb{Z}/p^2\mathbb{Z}$ given by $x \text{ mod } p\mapsto x \text{ mod }p^2$ for $0\leqslant x<p$.

Another bit of notation: let us write $X$ for the set of functions $C_p\to \mathbb{Z}/p\mathbb{Z}$, and $d:X\to Z^2(C_p,\mathbb{Z}/p\mathbb{Z})$ for the coboundary map, that is $$d(f): (x,y) \mapsto f(x) + f(y) - f(x,y).$$ By definition, $B^2(C_p,\mathbb{Z}/p\mathbb{Z})$ is the image of $d$.

Now we can identify the issue with what you are trying to do. What your teacher is telling you is that any $c\in Z^2(C_p,\mathbb{Z}/p\mathbb{Z})$ can be written as $$c= d(f) + t\cdot \chi$$ for a certain $f\in X$ and a certain $t\in \mathbb{Z}/p\mathbb{Z}$. But what you are trying to prove is that $c=t\cdot \chi$, which in general is just not true, even if you assume $c(0,0)=0$. What you seem to have understood from your teacher's advice is that $c$ is equivalent to some $c'$ with $c'(0,0)=0$, and that this $c'$ automatically satisfies extra properties (which actually amount to $c'=t\cdot \chi$). But in reality, that is incorrect: even though there are many $c'$ equivalent to $c$ such that $c'(0,0)=0$, only one of them is a multiple of $\chi$.

So now what you need to do is, given $c\in Z^2(C_p,\mathbb{Z}/p\mathbb{Z})$, to find $f\in X$ such that $c-d(f)$ is a multiple of $\chi$. (Note that this $f$ is not unique, unless you add an extra condition, like $f(1)=0$ for instance, but this is not very useful.) As your teacher points out, it's actually sufficient to show that $c-d(f)$ is $0$ on all $(x,y)$ with $x+y<p$ (which must clearly be true if it is a multiple of $\chi$). You should try to prove this on your own.

After that it just remains to find $f$ such that $c(x,y)=f(x)+f(y)-f(x+y)$ for all $x,y\in \{0,\dots,p-1\}$ such that $x+y<p$. As your teacher suggests, a natural method is to do that by induction on $x+y$, starting with $f(0)=c(0,0)$, and using the cocycle condition for the induction step.

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For a more theoretical flavor, note that what we are doing is essentially studying the exact sequence $$0\to B^2(C_p,\mathbb{Z}/p\mathbb{Z})\xrightarrow{d} Z^2(C_p,\mathbb{Z}/p\mathbb{Z})\to H^2(C_p,\mathbb{Z}/p\mathbb{Z})\to 0.$$

Indeed, we just showed that $Z^2(C_p,\mathbb{Z}/p\mathbb{Z})/B^2(C_p,\mathbb{Z}/p\mathbb{Z})$ is generated by a single element $\chi$, so $H^2(C_p,\mathbb{Z}/p\mathbb{Z})$ is cyclic. There is a natural identification $H^2(C_p,\mathbb{Z}/p\mathbb{Z})\simeq \mathbb{Z}/p\mathbb{Z}$, and the element $\theta(c)\in \mathbb{Z}/p\mathbb{Z}$ corresponding to the cohomology class $[c]\in H^2(C_p,\mathbb{Z}/p\mathbb{Z})$ is the only one satisfying $c=d(f)+\theta(c)\chi$ for some $f$.

You can check that $\theta(c)=\sum_{g\in C_p}c(g,1)$.