$\gcd(a,b,c)=\gcd(\gcd(a,b),c)\,$ [GCD Associative Law] Prove $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$ for $0\ne a,b,c\in \Bbb{Z}$. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference.
Hint $ $ By $\,\color{#c00}{\rm U}= $ gcd Universal Property (see below) we have
$$d\mid(a,b,c)\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid a,b,c\!\!\color{#c00}{\overset{\rm\ U}\iff} d\,\mid (a,b),c\!\!\color{#c00}{\overset{\rm\ U}\iff} d\mid ((a,b),c)\qquad$$
So $\,(a,b,c)\,$ &amp; $\,((a,b),c)\,$ divide each other (let $\,d\,$ equal them above) so they're equal, being $&gt; 0$.
This is the associative property of the GCD.  In the same way, by induction, we can erase (or normalize) brackets in $n$-argument gcds,  showing general associativity of the gcd, e.g. see here.
Remark $\ $ For completeness, below is a proof of $\,\color{#c00}{\rm U}= $ gcd Universal Property.
Lemma $\ \ d\mid a_1,\ldots,a_n\!\iff d\mid (a_1,\ldots,a_n)\ \ \ $ [GCD Universal Property]
${\bf Proof}\quad\ d\mid a_1,\ldots,a_n\Rightarrow\, d\mid (a_1,\ldots,a_n) = j_1 a_1\!+\ldots+j_n a_n\,$ for some $\, j_i\in\Bbb Z,\,$ by Bezout.
$\qquad\qquad\, d\mid (a_1,\ldots,a_n)\mid a_1,\ldots,a_n\,\Rightarrow\, d\mid a_1,\ldots,a_n\,$ by transitivity of $ $ &quot;divides&quot;,
The second divisibility in the prior line is due to the gcd being a common divisor of its arguments.
Remark $ $ Dually we have the universal property of LCM
Lemma $\ \ a_1,\ldots,a_n\mid m\iff {\rm lcm}(a_1,\ldots,a_n)\mid m\ \ \ $ [LCM Universal Property]
These universal properties are the definitions of GCD &amp; LCM in more general rings  - where the Bezout identity need not hold true,  e.g. in $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where the gcds $\,(x,2) = 1 = (x,y)\,$ cannot be written as linear combinations. Follow the links for further details.
Proof: Let $d=gcd(a,b,c)$ for positive integers $a$,$b$, and $c$. Then $d\mid a$, $d\mid b$, and $d\mid c$. Hence, $d\mid gcd(a,b)$ and $d\mid c$ and so $d\mid gcd(gcd(a,b),c)$. Thus, $d\mid f$ where $f=gcd(e,c)$ and $e=gcd(a,b)$. Since $f=gcd(e,c)$, then $f\mid e$ and $f\mid c$. Hence, $f\mid a$, $f\mid b$, and $f\mid c$, so $f\mid gcd(a,b,c)$ which means $f\mid d$. Since $d\mid f$ and $f\mid d$, $d=f$.