Hi @Ted

Waddup y'all

hi chat

If you asked a lot of people what the score of a handball match would be would the average of all the answers converge to the actual score of the match?

Handball isn't popular enough to perform any such analysis

basketball then

@user193319 I don't see how that wlog assumption holds. I might try to construct a counterexample.

@KarlKronenfeld Thanks for taking the time to think about this.

I've done a (hopefully correct!) proof of this before. I can see if it would help to send you it.

It's actually about something like this, but I had more assumptions, so it isn't likely to be helpful.

What is the gist of harmonic analysis

@user193319 The wlog assumption is valid in both (2.1) and (2.2). Write $[x,y]=x^{bp}$; this is possible since $[x,y]\in\langle x\rangle$ and $|x|=p^2$. Also, $b$ is invertible mod. $p$, because $[x,y]\ne 1$. Thus, there exists a multiplicative inverse $c$ of $b$ mod. $p$. Now, because $x$ and $y$ commute with $[x,y]$, we have $[x,y^c]=[x,y]^c$. Furthermore, $[x,y]^c=x^p$. Replacing $y$ with $y^c$, we get the desired equality.

Ah, I see. Very nice! Thanks!

Wait, do you mean $[x,y] \in \langle x^p \rangle$?

And you are saying that $Z(G) = \langle x^p \rangle$?

Yes, $Z(G)=\langle x^p\rangle$. I do not mean to say $[x,y]\in\langle x^p\rangle$, though it is true.

I could have made the second sentence clearer by also recalling that $[x,y]$ has order $p$.