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1926
bio website noldorin.com
location London, United Kingdom
age 24
visits member for 4 years
seen Jul 22 at 0:39

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

based in London, United Kingdom


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 suggested edit on An example of quasigroups with no identity
May
24
reviewed Approve suggested edit on How can I prove an ideal is a product of two irreducible ones
May
24
comment Proof of existence of primitive roots
That's true. I see what you mean... still not convinced there's a constructive proof, but let's see...
May
24
comment Proof of existence of primitive roots
Ah fair enough. :) I'm not sure myself how you could avoid a contradiction argument here, but I'll take your word for it!
May
24
comment Proof of existence of primitive roots
Interesting. I never thought of an approach this way. Thank you. I wonder if a proof other than by contradiction exists? Probably not, but oh well.