# Prove that $m=n$, if $\mathbb{Z}_m \cong \mathbb{Z}_n$

A question I got today was to explain why there is no subring of $\mathbb{Z}_{29}$ isomorphic to $\mathbb{Z}_8$.

First of all I looked for the subrings of $\mathbb{Z}_{29}$. Since the number of subrings of $\mathbb{Z}_n$ is the number of positive divisors of $n$, and $29$ is a prime number, the only subrings are $\{0\}$ and $\mathbb{Z}_{29}$.

In the notes it explains that for $\mathbb{Z}_m \cong \mathbb{Z}_n$, $m=n$. Therefore neither of the subrings can be isomorphic to $\mathbb{Z}_8$.

My question is, why must $m=n$? The notes do not explain this very well. Also, would my answer be correct using this?

Thank you

• Sorry I will change that now – user432533 Aug 2 '17 at 12:49
• Forget the algebraic structure and look at it set-theoretically, where the isomorphism is a bijection of sets. – Bill Dubuque Aug 2 '17 at 13:08

Since $\Bbb Z_m$ has $m$ elements and $\Bbb Z_n$ has $n$ elements, if there is a bijection between $\Bbb Z_m$ and $\Bbb Z_n$, then $m=n$.