Let $S$ be a subgroup of the group $\mathbb{Z}$. Then $S = \{ 0 \}$, else $S = \mathbb{Z}a$ with $a = \min\{ j \in S : j \in \mathbb{Z}^{+} \}.$ I was reading Group Theory and I got this theorem. This is my proof:

$\blacksquare~$Theorem: Let $S$ be a subgroup of the group $\mathbb{Z}$. Then

*

*Either $S$ is the trivial subgroup $\{ 0 \}$,

or else,

*

*it has the form $ \mathbb{Z} a $, where $a$ is the smallest positive integer in $S$.


$\blacksquare~$Proof: We can break the problem into two cases.

*

*Case I: When  $0 \in S $ and $\{0\}~$ is the only element of $S$ .


*Proof: We can easily solve the issue as $\{0\}$ is the trivial subgroup of any additive group.


*Case II: When $o( S ) < \infty $ but it contains other elements except $\{0\}$ too.


*Proof: Let  $n$ $\in$ $ S $ , then $ (- n) $ $\in$ $ S $.
Again, $ a \in S$ and $a$ is the smallest positive integer to be in $S$.
$\circ \circ~$ Let us prove at first, $\mathbb{Z}a$ $ \subseteq $ $ S $.
Then, we take an element $ n \in \mathbb{Z}a$. Therefore the element $ n $ has a form like $ n = ka $ for some $ k \in \mathbb{Z}$. Now, $ ka = \underbrace{a + a + a + \cdots + a}_k $.
Therefore, we have our result, by inducton, as $ a \in S$ $\Rightarrow$ $ ka \in S $ $\Rightarrow$ $ ( -ka ) \in  S.$
Therefore we have, $ \forall $ $ n \in \mathbb{Z}a $, $ n \in S $, from which we have shown, $\mathbb{Z}a$ $\subseteq$ $ S $.
$\circ \circ~$ Let us then show the reverse one, i.e. $ S $ $ \subseteq $ $\mathbb{Z}a$.
For this case let us take $ m \in S$. Therefore, by Euclidian Algorithm, we have the following ,
\begin{equation*}
    m = qa + r \quad [ 0 \leqslant r < a ]
\end{equation*}
Again, we know that, $\mathbb{Z}a$ $\subseteq$ $ S $. Then, we have, $qa \in S$.
Therefore , we obtain the follwing -
\begin{equation*}
    r = m - qa 
\end{equation*}
And  $ m , qa \in S $ . Then , we have that , $ ( m - qa ) \in S $ implying, $ r \in S $ contradicting the fact $ a \in S$ and $a$ is the smallest positive integer to be in $S$. Therefore $ r = 0 $, which tells us that, $ m = qa $.
Therefore we have proved that, $ S $ $ \subseteq $ $\mathbb{Z}a$ .
Therefore, we have from our two results,  $ S $ $ = $ $\mathbb{Z}a$ . And hence we are done!
We need to see the following example and understand some of our elementary number theory problems in light of group theory!

$\S .$ Extension of the Theorem:
$\bullet~$Proposition: Let $a , b$ be both integers and not both zero, and let the subgroup $S$ = $ \mathbb{Z}a + \mathbb{Z}b$ generated by $a$ and $b$ , and from previous theorem we know that, the subgroup $S $ can be represented as $\mathbb{Z}d $ where $d = \text{gcd} (a, b) $. Then-

*

*a) $d$ divides $ a  ~\&~  b $.


*b) If any $ e \in S $ divides both $ a ~\&~ b $ , then it must divide $ d $ also.


*c) The integer $ d $ , can be written as -
\begin{equation*}
    d = ra + sb 
\end{equation*}
for some $ r , s \in \mathbb{Z}$ [ Bezout's Theorem in Elementary Number Theory ].

$\bullet~$Proof: We see that, if $S$ is not the trivial subgroup of $\mathbb{Z}^{+}$ then, we can construct an additive group $\mathbb{Z}d$ generated by $d$ such that,
\begin{equation*}
    \mathbb{Z}d = \mathbb{Z}a + \mathbb{Z}b
\end{equation*}
Again we know that, $a \in S$ and $b \in S$ and $S = \mathbb{Z}d $. Which directly implies that, $d | a  $ and $d | b  \quad \quad  \cdots \cdots$ (a)
Again, if $ e \in S $, then, we have from (c)
\begin{align*}
    d = ka + sb,
\end{align*}
which implies $e \lvert d$.

Are there any glitches in the respective proofs?
Thanks
 A: You can do it faster. If $S$ isn't the trivial group (i.e. $S \neq \{ 0 \}$) then if $a = \min(S \cap \Bbb{Z}^+)$, clearly $S \supseteq \Bbb{Z}a$. (We know a nontrivial additive subgroup of the integers must contain a positive number because it is closed under additive inverses.) If there exists $b \in S, b \not \in \Bbb{Z}a$ then as you say, $S$ contains the positive number $c = \gcd(a, |b|) < a$, contradicting the fact that $a$ is the smallest positive element of $S$. QED.
A: A group presentation for $\Bbb Z$ is
$$\langle r\mid \varnothing\rangle.$$
This is because $\Bbb Z$ is cyclic, infinite, and generated by $1$, so all integer multiples (or, as in the presentation, all powers) of $1$ are in $\Bbb Z$ (or of $r$); presentations are usually multiplicative, but, here, by abuse of notation, we can consider $r=1$.
Let $S\le G$. Then $S\neq\varnothing$, since $0\in S$.
The trivial group is always (isomorphic to a copy of) a subgroup of any group, so let's suppose there is some $s\in S$ such that $s\neq 0$. Since $\Bbb Z$ is cyclic, $s$ is an integer multiple (or "power") of $r$, say, $s=rk$ for some $k\in \Bbb Z$. But the order of $r$ is infinite and, by closure of $S$, all multiples of $s$ are in $S$; and if some multiple $\ell$ of $s$ is the identity $0$, then $\ell=r(kt)$ for some $t\in\Bbb Z$, so that
$$\underbrace{r+\dots+r}_{kt \,\text{ times}}=0,$$
a contradiction.
Thus the order of $S$ cannot be finite (unless it is trivial).

Lemma: All subgroups of a cyclic group are cyclic.

Proof: (See here.) "$\square$".
Thus $S$ is infinite and cyclic. Hence $S\cong \Bbb Z$.
Such a subgroup $S$ of $\Bbb Z$ is necessarily, then, a group of all multiples of some multiple $s$ of $r$, using the notation above; that is, $$S=\Bbb Zs:=\{zs\mid z\in\Bbb Z\}$$ under addition. WLOG, as you have shown, we may assume $s>0$. Suppose, further, that $s$ is not the minimal positive multiple of $r$ in $S$, that, instead, there exists some $x\in S$ with $0<x<s$. But then, since $S=\Bbb Zs$, we have $x=ys$ for some $y\in\Bbb N$, which cannot be done as $0<ys<s$ implies $0<y<1$.
The result follows.

Your proof of the proposition is flawed, since $(c)$ is not proven.
