Showing Multiple of 4 in Finite Group Equation

[bjnartowt]

Homework Statement

In a finite group, show that the number of non-identity elements that satisfy the equation:

x^5 = e = identity element of multiplication mod n = 1

is a multiple of 4.


(Also need to show: if the stipulation that the group be finite is omitted, what can you say about the number of non-identity elements that satisfy x^5 = e ... but if I get the part of the question that's not in parenthesis, I should be able to figure it out...but I'm mentioning it just in case there's a hint in the asking of the question).

Homework Equations

axioms of groups: associativity, existence of inverse-elements, existence and uniqueness of identity, and we are in the (Abelian) group U(n), which is multiplication mod n. i think those are the ingrediants needed for a group.

The Attempt at a Solution

my professor suggested "look for a pattern", so i used Microsoft EXCEL to multiply mod n and find groups for U(n), for the cases n = 4, 5, 6, ... 14.

U(4) = \left\{ {1,3} \right\}
U(5) = \left\{ {1,2,3,4} \right\}
U(6) = \left\{ {1,5} \right\}
U(7) = \left\{ {1,2,3,4,5,6} \right\}
U(8) = \left\{ {1,3,5,7} \right\}
U(9) = \left\{ {1,2,4,5,7,8} \right\}
U(10) = \left\{ {1,3,7,9} \right\}
U(11) = \left\{ {1,2,3,4,5,6,7,8,9,10} \right\}
U(12) = \left\{ {1,5,7,11} \right\}
U(13) = \left\{ {1,2,3,4,5,6,7,8,9,10,11,12} \right\}
U(14) = \left\{ {1,3,5,9,11,13} \right\}

Then: of these elements, find only those guys that satisfy x^5 = e = 1. The elements of U(11): 3, 4, 5, and 9, four elements, satisfy x^5 = e. Everything else, only 1, the identity element (not a non-identity element! ha ha) satisfied x^5 = e.

Too small a set of instances (one!) to find a pattern. Is there some sort of closed form expression for 3, 4, 5, and 9 "in" U(11) that raising these guys to the 5th power that I could show to be mandatorily divisible by 4 that I am missing? Any suggestions on how to approach this? Am I going down a blind alley by just looking for a pattern? :-|

 

[bjnartowt]

I think I just "thought" of the instance where the group becomes infinite. if we were to stay in the integers, I think you can NOT have any solutions x^5 = e (if group is infinite, how can we speak of "mod"?). However, if we consider complex numbers, perhaps the solutions to x^5 = e must be the four (out of five) roots that are not equal to exp(i*2*pi) = the identity element.

 

[bjnartowt]

UPDATE: I have a nice proof that if x satisfies x^5 = e, then so too must x^2 and x^4.

Suppose there exists nonidentity element “x” that satisfies x^5 mod n = e. then:

\exists x:{x^5}\bmod n = e (1)

That means:

n|({x^5} - e) (2)

turn, n will divide any integer multiple of the right hand side of [I.15]. Such an integer-multiple might be:

({x^5} - e)({x^5} + e) = {x^{10}} - {e^2} = {x^{10}} - e (3)

Thus, if (1) is true and we thus have that “x”, then “x2” must also satisfy (1), generating our next “x”. By this same method, an “x4” must also exist, too: three out of the four elements that need to exist.

Does x3 also satisfy (1)? It must, somehow. Perhaps we could reach a contradiction if it didn't, and complete the proof?

 

[bjnartowt]

..at least i THINK i have a proof...critique this proof if you see problems! i have no experience with "pure-math" proofs!