Topics in Algebra I.N.Herstein Problem 7 Given that if $A$ and $B$ are cyclic of orders m and n and $\gcd(m,n)=1$ then $A\times B$ is cyclic.
Using this prove that if $u,v\in \mathbb Z$ then $\exists x$ such that $x\equiv u(\mod m);x\equiv v(\mod n)$
My try:
Now $(u,v)\in \mathbb Z\times \mathbb Z\implies (u\mod m,v\mod n)\in \mathbb Z_m\times \mathbb Z_n$  
By above lemma $\mathbb Z_m\times \mathbb Z_n$  is cyclic so $\exists (a,b)\in \mathbb Z_m\times \mathbb Z_n$  such that $(u\mod m,v\mod n)=x(a,b)$ for some $ x\in \mathbb Z\implies u\mod m=xa;v\mod n=xb$
I am getting stuck here .Any help

Consider the element $(1,1)$. What is the order of this element? It is the smallest $N$ that is both a multiple of $m$ and $n$. Since $m$ and $n$ are relatively prime, $N=mn$. This the group has $mn$ elements and an element of order $mn$, hence it is cyclic.
To show $x$ exists, note that $u,v$ determine an element in the product group, $(u,v)$. Since we have found a generator $(1,1)$, this means there exists an $x$ such that $x(1,1)=(x,x)=(u,v)$, hence $x\equiv u\pmod m$ and $x\equiv v\pmod n$.
