## Prove: order ab = ord a * ord b

Suppose (a,n) = (b,n) = 1 and gcd (ord a, ord b) = 1.
Show ord(ab) = ord(a)*ord(b).
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I think I just need a hint to carry this out. Thanks in advance.

$x = ord(ab) \iff (ab)^x \equiv 1 \mod{n} \iff a^x*b^x \equiv 1 \mod{n}$

since a and b are invertible mod n, then $a^x \equiv b^{-x} \mod{n}$

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