Order of an Integer mod m (number theory help)

[Ch1ronTL34]

Ok, my question is:

Show that if ab == 1 mod m, then
ord(m)a=ord(m)b

(Note that == means congruent) and ord(m)a means the order of a mod m

I know that if a^k==1 mod m, then the ord(m)a is the smallest integer k such that the congruence holds. For example,
ord(10)7=4 since 7^4==1 mod m.

I'm thinking, a^(ord(m)a)==1 mod m so maybe ab=a^ord(m)a and also:
b^(ord(m)b)==1 mod m so ab=b^ord(m)a

Am i on the right track? Thanks!

 

[matt grime]

I dislike unnecessary notation, so suppose the order of a is m, then what is 1^m=(ab)^m? Similarly if the order of b is n, what is 1^n=(ab)^n? Thus...? (remember the order is characterized by minimilality.)