Vtyjkyuk

Consider group $Bbb Z_pq$.
Consider the sets ${[p],[2p],cdots ,[(q-1)p]$ and ${[q],[2q],cdots ,[(p-1)q]$.

Show that there does not exist positive integers $r,s$ such that $r[a]=s[b]$ where $[a]in [p],[2p],cdots ,[(q-1)p]$ and $[b]in [q],[2q],cdots ,[(p-1)q]$.

Since $[a]in [p],[2p],cdots ,[(q-1)p]$ we have $[a]=[lp]$ where $1le lle q-1$ and $[b]=[kq]$ where $1le kle p-1$.

Assume that there exist positive integers $r,s$ such that $r[a]=s[b]implies [rlp]=[skq]$

But I don't understand how to get a contradiction from above?

Will someone please help.

From $[rlp]=[skq]$, we get that $rlp-skq equiv 0 pmodpq$. Said differently, we have
$rlp-skq=pqt$. This means $skq equiv 0 pmodp$. Assuming $p neq q$, we get $s equiv 0 pmodp$ (because $1 leq k <p$), likewise $r equiv 0 pmodq$. But then both $r[a]$ and $s[b]$ are the zero class $[0]_pq$.

Note: The way you have stated the question, we can have $r=q$ and $s=p$, in which case there does exist positive integers which will satisfy the given statement. However if we are looking for positive integers in the sense of modulo $pq$, then what I have stated above proves your statement.