finite and infinite question

[david]

find an example of an infinite group that has exactly two elements with order 4?

let R(5)=[1,7) is multiplication modulo 5, so there are infinite many elements within this interval, but only 2 and 3 have order 4.

2^1 mod 5=2
2^2 mod 5=4
2^3 mod 5=3
2^4 mod 5=1( this is the identity in multiplication)


3^1 mod 5= 3
3^2 mod 5= 4
3^3 mod 5= 2
3^3 mod 5= 1 ( again identity)

so |2|=|3|=4

do you think this is right?

 

[HallsofIvy]

So you are taking all real numbers between 1 and 7 "modulo 5"? That is, 5.5*6.2= 4.1 (mod 5)? How do you prove that there are no other numbers that are of order 4?

 

[david]

i'm not sure how to prove this, i've been testing numbers within this interval and mod 5 of any irrationals or rationals so far result in another rationals or irrationals or itself. in abstract algebra, multiplication modulo n means that taking an element m^n mod a such that the result is an identity.

 

[daon2]

The irrational numbers worthy of testing are only 4th roots of certain integers. a^4 = 5k+1 for an integer k if and only if a^4-1 is an integer divisible by 5...