Use Corollary 2 of lagrange's theorem to prove that the order u(n) is even when n>2

Question

Use Corollary 2 of lagrange's theorem to prove that the order U(n) is even when n>2.

Corollary 2: In a finite group, the order of each element of the group divides the order of the group.

Group U(n) is operation muiltiplication mod n. And, U(n)=｛1,2,3....n-1｝So, the order of u(n) is n-1. By Fermat's little theorem,For every prime p,a^p=a mod p. So,a^(n-1)= 1 mod n,so a^n= a mod n ?

But, I still don't know how to prove the order U(n) is even when n>2.How should I prove this?

score 3 · Answer 1 · answered Nov 14 '13 at 03:50

3

Hint: In order to show that the group's order is even, it suffices by Corollary to show that there's an element whose order is even. Can you think of a solution to the equation $$x^2 = 1$$ in your group such that $x$ is not $1$?

answered Nov 14 '13 at 03:50

Is it (n-1)^2= 1 mod n? – Deep Blue Nov 14 '13 at 04:00
@DeepBlue Yep. ${}{}$ – Nov 14 '13 at 04:01

score 1 · Answer 2 · answered Nov 16 '13 at 01:56

1

Proof: n-1 has order 2 in U(N),so 2 divides the order of U(N).

(n-1)^2=(n-1)(n-1)=n^2-2n+1= 1 mod n

So the order of U(N) is even.

answered Nov 16 '13 at 01:56

Deep Blue

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Use Corollary 2 of lagrange's theorem to prove that the order u(n) is even when n>2

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