Explain why there is no subring of $\mathbb{Z}_{19}$ which is isomorphic to $\mathbb{Z}_{6}$.

Question

This 2 mark question has been bothering me for a while because I think I know how its done but I am unsure how to explain it.

I am aware that if $\mathbb{Z}_{m} \cong \mathbb{Z}_{n}$ then $m = n$. Furthermore, since $19$ is a prime number then the only subrings of $\mathbb{Z}_{19}$ are $\{0\}$ and $\mathbb{Z}_{19}$.

Therefore we know that $\mathbb{Z}_{19}$ is not isomorphic to $\mathbb{Z}_{6}$. However, how do I show that $\{0\}$ is not isomorphic either?

Thank you

Answer 1

Every subring of $R=\mathbb{Z}/n$ is an ideal, see Show every subring of ring $\Bbb Z_n$ is ideal. Since for $n=19$, $R$ is a field, every ideal is $0$ or $R$. Both cannot be isomorphic to $\mathbb{Z}/6$ for cardinality reasons.

Answer 2

If $\Bbb Z_{19}$ had such a subring isomorphic to $\Bbb Z_6$, it's additive group would be a subgroup of the additive group of $\Bbb Z_{19}$.

By Lagrange, we would then have $6\mid 19$, contradiction.