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Q: Let the divisors of $p-1$ be $d_1,d_2,\cdots$Let $g \mod p$ be a primitive root ,then for each $d_i$ there is an element with period $d_i$

Mr. YI am trying,without success,to prove this statement: Let the divisors of $p-1$ be $d_1,d_2,\cdots$ Prove that if we have a primitive root $g \mod p$ ,then for each $d_i$ there is an element with period $d_i$ I am trying,without success,to prove this statement: Let the divisors of $p...

elementary-number-theory
Wojowu
Wojowu
What is $a$ and $t$? Also, $1^{1/q}$ need not be equal to $1$ modulo something.
Mr. Y
Mr. Y
Yes I've confused it all ,$t=p-1$ and $a$ is some integer which is not congruent to $0 \mod p$
Wojowu
Wojowu
In that case, $g=a^t$ never holds for $p>2$, because $a^t\equiv 1$ and $g\not\equiv 1$.
Mr. Y
Mr. Y
Can't I have $g=a^{t}$ for some real $t$ ?
Wojowu
Wojowu
But you said $t=p-1$.
Mr. Y
Mr. Y
14:02
Yes, I know but I am just confused.Would it work if $t$ is some real number ?(bear with me :S) also $q$ is an integer.
Wojowu
Wojowu
It isn't at all useful to consider real exponentiation for congruence problems, because it's not true that if $a\equiv b\mod p$, then $a^t=b^t\mod p$. Instead, I would suggest you looking at the integer powers of $g$ itself.
Mr. Y
Mr. Y
but if $b =1$ then it works,let me edit ,i think I can make it a little bit clear now.
Wojowu
Wojowu
Not necessarily. We have $4\equiv 1\mod 3$ but $4^{1/2}\not\equiv 1^{1/2}\mod 3$.
Mr. Y
Mr. Y
I guess my argument won't work even for $t$ integer?
Wojowu
Wojowu
Hello :)
Mr. Y
Mr. Y
14:02
Hello,thanks for your time
Wojowu
Wojowu
To answer your last comment
Mr. Y
Mr. Y
In my book $t$ is an integer
Wojowu
Wojowu
Even if $t$ is integer
There is a flaw, namely:
You first write $g=a^t$
But a line below you write $a^t\equiv 1$
Mr. Y
Mr. Y
My idea was to show that $d_i$ is some multiple of t which in turn is the period of some element
Wojowu
Wojowu
I'm not sure how that'd work
As I suggested in the other comment, try to look at powers of g
Mr. Y
Mr. Y
14:08
they must be divisors of p-1 ,also a^{p-1} \equiv 1
Wojowu
Wojowu
What must be a divisor of p-1?
Mr. Y
Mr. Y
the powers of g
Wojowu
Wojowu
No, they don't
For example, if we look modulo 5
And g=3
Then first power of g, which is 3, doesn't divide 5-1
Mr. Y
Mr. Y
Alright,I think I will just look for some proof if someone can provide it and study it.
Thanks for your time ;)
Wojowu
Wojowu
Andreas has provided a proof for you
In an answer
It's a result which will be useful for you, try to think about it
Mr. Y
Mr. Y
14:12
Hmm okay,thanks guys for your patience :P cya

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