To show that $7 \Bbb Z$ and $16 \Bbb Z$ are isomorphic as groups but not isomorphic as rings. To show that $7 \Bbb Z$ and $16 \Bbb Z$ are isomorphic as groups but not isomorphic as rings.
I have done the first part but finding difficult to show that they are not isomorphic as rings??
 A: For a group isomorphism, generator must go to generator (since the groups are cyclic).  Also any ring isomorphism has to be a group isomorphism of the underlying groups.
So for an isomorphism $\phi:7\mathbb{Z}\to 16\mathbb{Z}$, we must have that $7$ maps either to $16$ or $-16$.
Let's look at the case $\phi(7)=16$.  Then we should have $\phi(49)=\phi(7\times 7)=\phi(7)\phi(7)=16\cdot 16 = 256$.  But also $\phi(49)=\phi(7+7+\cdots+7)=\phi(7)+\phi(7)+\cdots \phi(7)=16+16+\cdots 16=112$.
These two are incompatible, so no such ring isomorphism exists. (The other possible isomorphism, with $7\mapsto -16$ is similar.)
A: $n \mathbf{Z}$ is generated by $n$, so that a group morphism $n \mathbf{Z} \to G$ is characterized by the value it takes on $n$. Therefore, the group morphism $7 \mathbf{Z} \to 16 \mathbf{Z}$ sending $6$ to $17$ is an isomorphism, and its inverse is defined by $16\mapsto 7$. Now take $f : 7 \mathbf{Z} \to 16 \mathbf{Z}$ a ring morphism. Then $f(7)k = 16$ for some $k$. But the $f(7) = 2^d$ for $0\leq d \leq 4$. Can you conclude from that ?
