Hi @BalarkaSen I have a question about what we talked about yesterday.
We have shown that $2^{2^6}\equiv 1 \mod{p}$ and that there is no smaller integer $y$ such that $2^y \equiv 1 \mod{p}$. Then $2^6$ is the order of $2$ in $\mathbb{Z}_p^{\ast}$. We also know that $2^{p-1}\equiv 1 \mod{p}$ and so we deduce that $2^6 \mid p-1$. Is this right? We don't have to use lagrange, do we?
If $x,y$ are rational and $y^2=x^3-2x$ then there are integers $a,b$ such that $t_1=(a^2+2b^2)^2$,$t_2=4ab|a^2-2b^2|$, $t=\frac{|t_1^2-2t_2^2|}{t_1t_2}$, $x=\frac{t\pm\sqrt{t^2+8}}2$
If a, b, c, ... are elements of a group G then <a, b, c, ...> is in general a subgroup of G - definitionally so, because it's the "subgroup generated by a, b, c, ..."
I liked the idea of flattening the knot, but the calculations needed to use SvK don't look too bad, I might do that after this complex analysis lecture
In particular: $G$ a group, $\varphi$ an automorphism, then $x \sim y \iff \exists z: x = zy\varphi(z)^{-1}$ is an equivalence relation. How many equivalence classes are there?
It's connected to Lefschetz and Nielsen fixed point theory :)
The groups I work on are almost-crystallographic groups $\Gamma$. Say, if $G$ is a connected, simply connected nilpotent Lie group, then define the affine group $\operatorname{Aff}(G) = G \rtimes \operatorname{Aut}(G)$. If $C$ is a maximal compact subgroup of the affine group, then $\Gamma$ is a cocompact discrete subgroup of $G \rtimes G$.
I guess, if you want an "interesting" result, there's this thing my promotor did. It was already known that every finitely generated abelian group had an automorphism with finitely many equivalence classes.
He proved that there exists uncountably many infinite countable abelian groups (non-finitely generated) which do not have such automorphism
And the link with topology, is something along the lines of "take a sufficiently nice space and a self-map. Count the number of equivalence classes of the induced automorphism of the fundamental group; then this number is a lower bound for the number of fixed points of the self-map"
Let $p \in \mathbb{C}$ and $f$ a complex function. If there is some $U \ni p$ such that $f$ is holomorphic on $U\setminus \{p\}$ and there exists some holomorphic function $g$ on $U$ with $g(p) \neq 0$ such that $f(z) = g(z)/(z-p)^n$ on $U\setminus \{p\}$ for some positive integer $n$, then $f$ has a pole of order $n$ at $p$
so I have Calculus on Manifolds and Algebra 2 classes from 1 pm to 3 pm, each class being an hour, and this is actually lunch time
I otherwise nap after comfortably having my lunch, but because of these classes, I have to hurriedly eat my lunch only to go to class and find myself too sleepy to focus
I might as well not attend these classes because I can't remember so many things the profs talk about in these classes
You could take circle with centre $(0,0)$ and radius $r$, and a point $P = (0,-r)$. Then any other point $Q$ is of the form $Q = (r \cos \theta, r \sin \theta)$.
Pythagoras then quickly gives the distance between $P$ and $Q$ to be $r \sqrt{2(1-\sin \theta)}$, I think
