Suppose ka = kb (mod n). Prove that if gcd (k, n) = 1, then a = b (mod n).?

Notation: I'm using
≡ for "is congruent to" and
| for "divides."

If ka ≡ kb (mod n), then for some positive integers i, j, and r,
ka = in + r and
kb = jn + r

Therefore,
ka - kb = in + r - (jn + r)
k (a - b) = (i - j)n

So n | k(a-b)

Therefore, if gcd(k,n) = 1,
n | (a-b) and
a ≡ b (mod n)