Sub-group of the modulo group is an ideal

Question

how do I show that every sub-group of ring $Z_n$ is an ideal in $Z_n$?

If $n$ is prime, the only sub-groups are the trivial and that mean's they are ideals, but if $n$ isn't prime, there are non trivial groups.

My second thought is that every sub-group is $mZ_n$ for $0<m<n$ and then it isn't hard to show that it is an ideal, but is it correct and how do I prove it generally (that there isn't any ideal from another form) ?

Answer 1

This follows from the observations below:

• Every subgroup of $\mathbb Z/n\mathbb Z $ corresponds to a subgroup of $\mathbb Z$ that contains $n \mathbb Z$.

• The subgroups of $\mathbb Z$ that contain $n \mathbb Z$ are precisely $d\mathbb Z$, with $d$ a divisor of $n$.

• Every subgroup of $\mathbb Z$ is an ideal.

• The canonical group homomorphism $\mathbb Z \to \mathbb Z/n\mathbb Z$ is actually a ring homomorphism.

Answer 2

An ideal has to be a subgroup (of the additive group of the ring) to begin with. So if you prove that all subgroups are already ideals, you are done.

PS Just correct the statement

every sub-group is $mZ_n$ for $0<m<n$

to

every subgroup is of the form $m Z_{n}$ for $m \mid n$.