Let p and q be distinct prime numbers and let n = pq. Show that $\displaystyle HK = \mathbb{Z}^x_n $ for the subgroups $\displaystyle H = \{[x] \in \mathbb{Z}^x_n | x \equiv 1 (mod \ p)\} $ and $\displaystyle K = \{[y] \in \mathbb{Z}^x_n | y \equiv 1 (mod \ q)\} $.

Hint: use the chinese remainder theorem to show that the sets are the same.

so i have that any element of HK is of the form xy (mod pq) and i also know that gcd(p, q) = 1 which hints to me that i can use the chinese remainder theorem somehow. i have that $\displaystyle xy \equiv y (mod \ p) $ and $\displaystyle xy \equiv x (mod \ q) $ which doesn't seem to help me much. then i tried saying that if $\displaystyle y \equiv 1 (mod p) $ and $\displaystyle x \equiv 1 (mod q) $ then by CRT there exists a unique z such that z satisfies those congruences. if z = xy then by CRT $\displaystyle xy \equiv 1 (mod pq) => xy \equiv 1 (mod n) $ so the element xy is in the set of units modulo n. I don't know if my reasoning was correct though. i "forced" y to be congruent to 1 (mod p) but since it was already congruent to 1 (mod q) it seems that y must be in other set and same for the x's so i essentially made H = K i think. i also have to prove the reverse inclusion which is that the set of units modulo n is contained in HK which i do not know how to proceed with. can someone help me with the "forward" and "backwards" directions of this proof? or if i don't need to consider the forward and backwards cases how would i proceed? thanks.