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## Main Question or Discussion Point

Please help me prove this. I worked hard to make my notation somewhat easy to follow:

If gcd ( ord_n(a) , ord_n(b) ) = 1 , then ord_n(a*b) = ord_n(a) * ord_n(b)

Attempted proof:

I can't see how to use the gcd condition. Which is bad news. I do realize the following.

Let k1 = ord_n(a) and k2 = ord_n(b). Hence, a^k1 == 1 (mod n) and b^k2 == 1 (mod n).

Thus, (a^k1)*(b^k2) = = 1 (mod n).

I am very stuck. Can you please help me?

HFO8

If gcd ( ord_n(a) , ord_n(b) ) = 1 , then ord_n(a*b) = ord_n(a) * ord_n(b)

Attempted proof:

I can't see how to use the gcd condition. Which is bad news. I do realize the following.

Let k1 = ord_n(a) and k2 = ord_n(b). Hence, a^k1 == 1 (mod n) and b^k2 == 1 (mod n).

Thus, (a^k1)*(b^k2) = = 1 (mod n).

I am very stuck. Can you please help me?

HFO8