gcd and lcm - Suppose $gcd(a,y)=s$ and $gcd(b,y)=t$. Prove that $gcd(gcd(a,b),y)=gcd(s,t)$.

All I have so far is that

$$s|a, s|y, t|b, \text{ and } t|y.$$ I also know

$$\gcd(\gcd(a,b),y)=\gcd(a,b,y)=\gcd(a,gcd(b,y))$$

by the associative property of gcd. It would suffice to show

$$\gcd(a,b,y)=\gcd(gcd(a,y),\gcd(b,y)).$$
I'm just not sure how to prove it. Thanks for your help.

Answer

I would approach it a bit differently. Let $d=\gcd(\gcd(a,b),y)$. Then $d\mid\gcd(a,b)$, and $d\mid y$. Since $d\mid\gcd(a,b)$, we also know that $d\mid a$ and $d\mid b$. Since $d\mid a$ and $d\mid y$, we know that $d\mid s$; similarly, $d\mid t$, so $d\mid\gcd(s,t)$.

Now let $e=\gcd(s,t)$ and make a similar argument to show that $e\mid d$. Since $d,e\ge 1$, $d\mid e$, and $e\mid d$, it must be the case that $d=e$.