# Prove: order ab = ord a * ord b

• Nov 20th 2010, 04:10 PM
$x = ord(ab) \iff (ab)^x \equiv 1 \mod{n} \iff a^x*b^x \equiv 1 \mod{n}$
since a and b are invertible mod n, then $a^x \equiv b^{-x} \mod{n}$