GROUP THEORY DUMB QUESTION
- Home
- Board: /sci/ - Science & Math
- Reading: GROUP THEORY DUMB QUESTION
Let [math] \sigma_n : ( \mathbb{Z} / 2009 \mathbb{Z} )^{*} \rightarrow ( \mathbb{Z} / 2009 \mathbb{Z} )^{*} [/math] be defined by [math] x \mapsto x^n [/math]. Find the smallest [math] n > 1 [/math] such that [math] \sigma_n [/math] is a bijection.
Please help me.
>>8154250
n =1
>>8154258
Larger than one ya dingus, it's right there in the question.
>>8154250
n=9
>>8154250
n=2
>>8154250
2009 isn't prime, and 287^2 = 0 mod 2009. It's impossible OP.
>>8154250
n=
>>8154344
Wh-what?
let n be the number you want
by slanty eyed remainder theorem, (Z/2009Z)*=(Z/49Z)* x (Z/41Z)* =(Z/42Z)x(Z/40Z)=(Z/2Z)x(Z/3Z)x(Z/7Z)x(Z/5Z)x(Z/8Z)
by sylow's theorem there exists an element of order 2,3,5,7 in (Z/2009Z)* so the n^th power map with n=2,3,5,7 can't be injective
the next smallest prime is 11
if the n-th power map with n=11 wasn't injective there would be an element of order 11, but 11 doesn't divide #(Z/2Z)x(Z/3Z)x(Z/7Z)x(Z/5Z)x(Z/8Z)=1680 (which it would need to by lagrange's theorem)
so n=11
wolfram also confirms x^11 = 1 mod 2009 has only x=1 as a solution
>>8154409
nice anon!
>>8154383
you know that's the answer.
>>8154409
Awesome dude!
>>8154423
A friend of mine had that answer but wouldn't tell me how he got it.
