Day 33
11:40 pm, 14th may 2026
I have an EF interview tmrw at 4pm.
Let's just make every single day so fucking crazy
let's take a quick caffeine nap, i'll tell kuku to wake me up. lol i rmbr he used to wake me up to study in lko, during my jee time, and now he has jee in 3 days. life moves so fast, we're gonna make the most of it every single day.
9:32 am, 15th may 2026
lmao the caffeince nap turned into full sleep. lemme do some more math rn.
I'm trying to find all groups of order $n$ for all n i can.
So i guess we just need to fill out the whole multiplication table under the group rules. $n=1,2,3,4$ i just did. ez. $n=p$ is ez too. something must have order $p$ and then $\{1,x,\cdots x^{p-1}\} \simeq \mathbb{Z}/p\mathbb{Z}$
for $n=9$ there are only two groups. pretty cool proof i found.
9:57 am
Lol I proved there are only two groups of order $p^2$. kinda genius lmao.
so if anything has order $p^2$ you're done. else every element has order $p$.
Then consider $\lt x \gt = \{ x, x^2 , \cdots x^{p-1} \}$
choose $a_1$ randomly. choose $a_2$ not in $ \lt a_1 \gt$. choose $a_3$ not in $\lt a_1 \gt$ or $\lt a_2 \gt$
also note a simple thing that $$a_x \notin \lt a_y \gt \iff a_y \notin \langle a_x \rangle$$
do this till $a_{p+1}$
now you're still not done at all, cos the multiplication table is not defined, but my key insight was you can infact define
$a_1,a_2$ and then $a_n=a_1^{n-2}a_2$ and this works. so only possible group is hence $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$
(you can do this by just taking $a_1=(1,0)$ and $a_2=(0,1)$.
damn that's kinda genius lmao. like completely vibes based proof of this fact haha.
10:17 am
$\mathbb{Z}/pq\mathbb{Z} \simeq \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/q\mathbb{Z}$ cos $(1,1)$ has order $pq$!!!
10:33 am
found all groups of order $2p$.
ok so either someething has order $2p$ and you're done. if anything has order $p$, consider $\langle x \rangle$
then the rest $p$ elements must have order $2$, infact just consider $a \notin \langle x \rangle$. then, $a,ax,ax^2 \cdots ax^{p-1}$ are the $p$ elements with order 2. (note $a \neq x^{\alpha}$ now raise both sides to power $\alpha ^ {-1} (\text{mod p})$ and you see $a^{\beta} \neq x$ )
This is actually equivalent to $D_{2p}$ with $a=s$ and $x=r$. and obv all elements cant have order 2. uhhh wait why not, why cant you have $2p-1$ elements of order 2? fuck i didnt think of this.
3:23 pm
i'd gone for a haircut and played ball w/ kuku...
5:44 am 17th may 2026
Man, im just gonna do so much cool shit every sinlge day man. im working every single moment now, literally nothing else matters. witness history.
let's have some fun now.
thinking about why you can't have $2p-1$ elements of order 2.
6:24 am
ok damn this in general, theory of groups is really really interesting, especially interested in the computational stuff. lets discover evreything ourselves...
4:24 pm
Kuku is giving JEE Adv right now. i love you so much bro .
4:44 am, 19th May 2026
I am going to achieve all of my dreams. do my absolute best every single day. i dont give a fuck now. im locked in as fuck. let's get this shit done man , every sinlge day now. every single day. every single moment. not wasting any time now. new blogs every single day, lets have fun .