2,758 reputation
1926
bio website noldorin.com
location London, United Kingdom
age 24
visits member for 4 years, 5 months
seen Dec 8 at 22:50

entrepreneur; graduate in mathematics / theoretical computer science / theoretical physics; polymath-in-training

based in London, United Kingdom


Sep
26
comment Is Aluffi's “Algebra. Chapter 0” a good introduction to algebra?
Not particularly challenging?! The exercises in Aluffi are well-known to be incredibly difficult.
Sep
11
comment Is Aluffi's “Algebra. Chapter 0” a good introduction to algebra?
Care to elaborate which you prefer?
Aug
7
comment A finite abelian group whose order is divisible by 10 contains an element of order 10
Okay, thanks. Unfortunately I have a few similar question to the asker but do not have the FTAG yet however.
Aug
5
comment A finite abelian group whose order is divisible by 10 contains an element of order 10
How on earth does this help? Surely not all abelian groups are isomorphic to $Z_n$? Or perhaps I just haven't covered this yet in my book...
Jul
20
awarded  Yearling
Jul
7
awarded  Pundit
Jul
2
awarded  Curious
May
25
comment Proof of existence of primitive roots
Haha, yes. I completely messed up there... I could have started with $1 \equiv a^{p-1} \equiv a^{nd}$, but I don't think that gets us anywhere... eh.
May
25
comment Proof of existence of primitive roots
Not sure what I was thinking any more, in fact. I believe it fails. :)
May
25
comment Proof of existence of primitive roots
Oh yes. $a$ is just an arbitrary element of the group, the one with order denoted $k$.
May
25
comment Proof of existence of primitive roots
As it happens, your proof/explanation gave me ideas, and I think I've come up with a really simple proof now. Perhaps you could kindly confirm? Using notation in my original question, suppose $k \nmid n$. Then $n = m k + r$, where $0 < r < k$. Now $1 \equiv a^n \equiv a^{mk} a^r \equiv (a^k)^m a^r \equiv a^r $, which contradicts that the order of $a$ is $k$.
May
25
comment Proof of existence of primitive roots
Thank you, all looks good now!
May
25
revised Proof of existence of primitive roots
deleted 4 characters in body
May
25
comment If $\gcd(a,b)=1$ and $a$ and $b$ divide $c$, then so does $ab$
Easily the better answer here.
May
25
accepted Proof of existence of primitive roots
May
25
comment Proof of existence of primitive roots
The website is suggesting we move this discussion to chat, but I'm happy to continue it wherever, FYI. :)
May
25
comment Proof of existence of primitive roots
Hah. I know nothing of the life of Banach I'm afraid, but that sounds wise to me! And thank you.
May
25
comment Proof of existence of primitive roots
It's the latter. But no worries, take your time, and enjoy your Saturday night. ;)
May
25
comment Proof of existence of primitive roots
I'm afraid that on rereading your proof, I don't quite see how the last sentence follows. I can see how one constructs an element of order that divides $m'n'$, but not of that exact order.
May
24
reviewed Approve An example of quasigroups with no identity