Proof that $n\Bbb Z \leq \Bbb Z$ and are the only subgroups of $\Bbb Z$

My challenge is

Prove that if $n = 0,1,2,\ldots$ and $n\Bbb Z = \lbrace nk: k \in \Bbb Z \rbrace$, show that $n\Bbb Z$ is a subgroup of $\Bbb Z$ and are the only subgroups.

I handled the first point by saying that

i) Identity is 0, and $0\in n\Bbb Z$ and $0 \in \Bbb Z$

ii) For any $a,b \in n\Bbb Z$ $a+b = na_2 + nb_2 = n(a_2+b_2) \in n \Bbb Z$

iii) For any $na \in n\Bbb Z$, its inverse $n(-a) \in n\Bbb Z$.

For the second point, I said that all of the subgroups should be infinite, because for any finite group the largest element $a+a \notin H$, where $H$ is the potential subgroup, which is a contradiction. So $H$ must be of infinite order. For any $a \in H, -a$ should also be in $H$. The difference between any $a,b \in H$ should also be in $H$, i.e. $(a-b) \in H$. This leaves only the whole group or any group of the form $n\Bbb Z$.

Is my proof complete?

Solution 1:

Hint: A simple way to tackle this is by considering $$n = \min \{x \in H ; x> 0\}$$

where $H \leq \mathbb Z$.

The inclusion $H \supseteq n\mathbb Z$ is clear.

For the other one take any $h \in H$ and use the division algorithm and the fact that $n$ is minimum.

Solution 2:

You're almost there but not quite. You've shown that any subgroup of $\Bbb Z$ has a copy of $n\Bbb Z$. The way you want to go about it is to consider the least positive element in the subgroup $H$ and show that element generates $H$. (You'll end up doing a proof by contradiction.)