Find $e$ in set closed under subtraction, hence prove co-prime elements have GCD $=1$.

Question

In the book by Andre Weil, titled : Number theory for Beginners, is given on page 4,5 theorem #II.1 regarding a set M i.e. closed under subtraction. There are also two corollaries to the same, followed by another theorem #III.1.

I think the closest other post (found only one) on the same topic are: a, that too was in search of something like the book here states (please correct me if wrong).

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Theorem II.1. Let $M$ be a non-empty set of integers, closed under subtraction. Then there exists a unique $m>0$ such that $M$ is the set of all multiples of $m: M=\{mz\}_{z\in Z} = mZ$.

Proof. If $x \in M$, then, by assumption, $0 = x - x \in M$ and $- x = 0 -x\in M$. If also $y \in M$, then $y+x = y -(-x) \in M$, so that $M$ is also closed under addition. If $x \in M$ and $nx \in M$, where $n$ is any positive integer, then $(n + 1)x = nx + x\in M$, therefore, by induction, $nx\in M$ for all $ n > 0$, hence also for all $n\in \mathbb{Z}$.
Finally, all linear combinations of elements of $M$ with integral coefficients are in $M$; as this property of $M$ obviously implies that $M$ is closed under addition and subtraction, it is equivalent with our assumption on $M$.
If $M=\{0\}$, the theorem is true with $m = 0$. If not, the set of elements $>0$ in $M$ cannot be empty; take for $m$ the smallest one.
All multiples of $m$ are then in $M$. For any $x\in M$, apply the lemma and write $x=my + r$ with $0\leq r \lt m$; then $r = x - my$ is in $M$. In view of the definition of m, this implies $r = 0, x = my$. Therefore $M=mZ$.
Conversely, since $m$ is the smallest element $\gt 0$ in $mZ$, it is uniquely determined when $M$ is given.

Corollary 1. Let $a,b,...,c$ be integers in any {finite) number. Then there is a unique integer $d\ge 0$ such that the set of all linear combinations $ax + by+... + cz$ of $a,b,...,c$ with integral coefficients $x,y,... ,z$ is the set of all multiples of $d$.

Proof. Apply theorem II. 1 to that set.

Corollary 2. Assumptions and notations being the same as in corollary 1, $d$ is a divisor of each one of the integers $a,b,...,c$, and every common divisor of these integers is a divisor of $d$.

Proof. Each one of the integers $a,b,...,c$ belongs to the set of their linear combinations. Conversely, every common divisor of $a, b,..., c$ is a divisor of every one of their linear combinations, hence in particular of $d$.

Definition. The integer $d$ defined in the corollaries of theorem II.1 is called the greatest common divisor (or in short the g.c.d.) of $a, b,..., c$; it is denoted by $(a,b,...,c)$.

As the g.c.d. $(a, b,..., c)$ belongs to the set of linear combinations of $a, b,..., c$ (since it is the smallest element $\gt 0 $ of that set, unless $a, b,..., c$ are all $0$), it can be written in the form $$(a,b,...,c) = ax_0 + by_0 + \cdots +cz_0 $$ where $x_0,y_0,...,z_0$ are all integers.

Theorem III.1. Integers $a,b,...,c$ are mutually relatively prime if and only if the equation $ax + by+ ... +cz=1$ has a solution in integers $x,y,...,z$. In fact, if $(a,b,...,c)= 1$, then, by corollary 1 of theorem II.1, the equation in question has a solution.

Conversely, if it has a solution, then every common divisor $d>0$ of $a,b,...,c$ must divide $1$ and must therefore be $1$.

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Beginner like me has complaint that the book is not articulate, nor states examples (nor implores in exercises) about such $m$, for some set $M$, apart from other implicit details as stated below.

Theorem II.1 gets $m$ as "identity" element, for the set $M$. Also, proves it to be the smallest element in the set. Also, shows all elements in the set are multiples of it.

So, $d$ is same as $m$.

Also, $m$ can be $0$, only for empty set $M =\{0\}$.

But, $ax + by +...+ cz = nd$ with $x, y, z, n$ as integers (positive, negative, or zero).

If so, then for positive $n$; $nd$ for $d$ as identity element means that $ax + by +...+ cz = nd \implies n(a'x + b'y +...+ c'z = d)$

But, if $d =$ unit (identity) element in set $M$, then $n$ is never a negative value, as GCD (here, of co-prime values: $a,b,...,c$) is always $1$. So, $n$ cannot be negative. This is in turn directly based on identity element$d$ being non-negative, and for non-empty set as non-zero.

As per author, it is possible in Corollary 1 that if $d=0$, then $n$ can be $ = 0$, or a nonzero value.

But, it is not clear when the linear combination $ax + by +...+ cz = 0$, as for me it is known that linear combination of co-primes is $n.d = n.1$. Else, in linear equalities it is common to see the sum as $=0$. But, it seems quite different to have all integers in set (in complete ordering, I suppose; i.e. no gap from $d=e$ to largest integer in the set $M$) involved in linear combination.

So, want to ask significance of $ax + by +...+ cz = 0$.

The linear combinations are used based on closure wrt addition & subtraction operations in the set $M$.

Theorem III.1 is simpler as is stated with co-prime elements, $a,b,...,c$, with $d=1$.