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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!