Prove that if $\{7k:k\in{\mathbb Z}\}\subsetneq\{nm:m\in{\mathbb Z}\}$, then $n=1.$

Question

Let n be a natural number. Prove that if $\{7k:k\in{\mathbb Z}\}\subsetneq\{nm:m\in{\mathbb Z}\}$, then $n=1$.

I know that we must show $x\in{A}$ implies $x\in{B}$, and that there exists $x\in{B}$ such that $x\notin{A}$, which means that $x\in{A}\neq{x\in{B}}$. Any help is appreciated; thanks

Answer 1

The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n \neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.

Answer 2

Hint $\ $ Show $\ a\Bbb Z\supseteq b\Bbb Z\iff a\mid b.\ $ Hence $\,n\Bbb Z\supseteq 7\Bbb Z\iff n\mid 7$