I have 2 related problems: Every subgroup of $\mathbb{Z}$ (under addition) is a subring (and an ideal), and the same thing but for $\mathbb{Z_n}$ (I'm following the definition for rings that doesn't require multiplicative identity).
Intuitively, I see why this is the case, since every subgroup of $\mathbb{Z}$ is just the multiples of some integer and if you multiply that by any other integer you get to another multiple of it, and it's similar for $\mathbb{Z_n}$. However, I'm having a hard time figuring out what a valid proof of this looks like, I saw some question about this with an answer saying you could prove it by induction and I'd like to see how that would work, but also would appreciate other methods.
I've been trying to prove it "rigorously" but still find working with rings a bit unwieldy and get confused on what is valid and what is not since here you can see multiplication as multiplication by some element of the ring, but since they are integers also as repeated addition. I know this has been asked but I did not find the previous answers clear enough on the whole proof process.
Best Answer
Let's prove that every subgroup $A$ of $\mathbb{Z}$ is an ideal. We need to show:
The first three properties are just the assumption that $A$ is a subgroup. For the last property note that $na = \underbrace{a + a + \dots + a}_{\text{$n$ times}}$, which lies in $A$ because of property 3 / because $A$ is a subgroup. Actually, my last sentence only really works when $n \geq 0$, but for $n < 0$ just use $na = -(-na)$ and combine the argument for non-negative $n$ with property 2.
That it's also a subring depends on the definition of subring. Usually people require subrings to contain 1, so that the statement would be false. If they don't require it, then we need to prove:
But that follows from property 4 since $A = \mathbb{Z}$.
The case $\mathbb{Z}/n\mathbb{Z}$ is similar.