According to theorem if $H$ is a subgroup of index $m$ of a free Group $G$ of rank $n$, then $H$ is free and rank of $H$ is $nm+1-m$.
Now if $G= Z\times Z$ then it is free group of rank 2 if $H= Z\times 2Z$ then $H$ is of index $2$ in G, then by theorem $H$ has rank $3$
but I know generators of $Z\times 2Z$ are $(1,0)$ and $(0,2)$
What is wrong with me?
Although the groups are free abelian groups, they are not free groups as the theorem requires.