Multiplicative order

 If gcd(a,n) = 1, then:

for some positive integer 'k'
a^k = 1 mod n

Why?
Why is it certain that for some k we will get a^k = 1 mod n?
WLOG, let there be i > j s.t. a^i = a^j mod n.
There will be such i,j for sure because there are infinitely many powers but remainders have only 'n' possible values.
=> a^i - a^j = c.n (0 mod n)
a^j ( a^(i-j) - 1) = c.n
Since a,n are coprime => n divides a^(i-j) - 1 => we found a power of k s.t. a^k = 1 mod 'n'.

Now,
Multiplicative order of 'a' modulo 'n' is the smallest 'k' s.t. a^k = 1 mod 'n'.
=>
if a^m = 1 mod n then k | m
Why?
Let m = xk + y by division algorithm. So 0 <= y < k
a^(xk + y) = 1 mod n = a^xk.a^y = a^y = 1 mod n
But y < k and we had said that k was the smallest integer for which a^(something) = 1.
Hence proved by contradiction.

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