@AlexanderGruber Actually, I have seen it somewhere, as I have said. When Pedro said something about generalized quarternions, I googled what was it, so it was half-revealed already.
@BalarkaSen it is. but many people are hesitant to admit when they don't understand something, because trying to sound smarter than everyone else is a widespread practice in academia, which makes people defensive
Let $G$ be a group of order $2^n$ with generators $a,b$ satisfying $a^{2^{n-2}}=b^2$ and $b^{-1}ab=a^{-1}$. Since $b^{-k}ab^k=a^{-k}$, $b$ cannot have order $2$; because it gives by $k=2^{n-2}$ ($n\geqslant 3$) that $a=1$, which is impossible by $|G|=2^n$. Since $a^{2^n}=b^8=1$; then $|b|=4,8$. But it cannot be $8$ because then $|a|=2^n$ and $G$ has too many elements. Then $|a|=2^{n-1}$ and $|b|=4$.
Let $\mathbb{H}$ be the ring of Hamiltonian quaternions - that is, a real space with basis $1,i,j,k$ satisfying $$i^2=j^2=k^2=-1\hspace{50pt}i j=- j i=k,jk=-kj=i,ki=-ik=j.$$
take $V=\langle a,b|a^3=b^3=[a,b]=1\rangle$, and take the following elementary abelian groups: $X=\langle x_1,x_2,x_3 | x_1^7=x_2^7=x_3^7=[x_i,x_j]=1\rangle$ and $Y=\langle y_1,y_2,y_3 | y_1^{13}=y_2^{13}=y_3^{13}=[y_i,y_j]=1\rangle$. Form $(X\times Y)\rtimes V$ by letting $x_i^a=x_{i+1}$, $x_i^b=x_i^3$, $y_i^b=y_i^2$, $y_i^a=y_{i+1}$, where the indices are taken mod $3$
is it true that $(X\times Y )\rtimes V$ has no element of order $3\times 7 \times 13 =273$?
@PedroTamaroff it's super significant, i use it all the time. but its proof and the way it's used aren't very similar.
so, $a$ is acting on $X$ by cyclic permutation of the basis elements and $b$ is acting on $X$ by raising everything to a power such that that the automorphism $x\mapsto x^b$ of $X$ has order $7$
and same thing for $Y$ except their roles are reversed (and using $13$ instead of $7$)
I have been exposed to limit inferior and limit superior, yes. So as delta shrinks we are reconsidering each supremum $f$ takes on these delta intervals about y?
Ahh, I thought it was over all $\delta>0$ which is why I was confused. I thought to myself, Wouldn't this value just be the maximum on the domain if we consider all $\delta >0$?
Just thinking what's going on in Ukraine/Russia, it occurs to me that I've never known of someone from that part of the world to show up here. It could be that all people of Russian origin are math studs and don't need to ...
If $E$ is an elliptic curve with point at infinity $\mathcal{O}$ and $P$ is a point on $E$ of order $m$, a rational function $f_P$ is a rational function with he formal sum $m[P] - m[\mathcal{O}]$. I don't understand how to evaluate $f_P(S)$ for some point $S$ on $E$. Anyone able to explain this?
