Properties of Greatest Common Divisors I really want to have help verify these properties of GCD:
Let $s,t \in \mathbb{Z}$ and $m,n$ be positive integers with $m\mid n$.


*

*If $\gcd(t,n)\mid\gcd(s,n)$, then $\gcd(t,m)\mid\gcd(s,m)$.

*If $\gcd(t,n)>\gcd(s,n)$, then $\gcd(t,m)\geq\gcd(s,m)$.


For 1, I know that $\gcd(t,m)\mid\gcd(t,n)$ and so by transitivity, $\gcd(t,m)\mid\gcd(s,n)$. Also, $\gcd(s,m)\mid\gcd(s,n)$ and so $\gcd(t,m)\cdot k = \gcd(s,m) \cdot t$ for some $k,t \in \mathbb{Z}$. I just really want to verify whether $\frac{k}{t}$ is an integer so that the conclusion follows.
For, 2, it is clear that the less than relation follows, but I can't understand the equality condition.
I hope someone will help me... Thank you!
 A: For 1, let $x = \gcd(t, m)$. Clearly $x$ divides $m$. You have to prove that $x$ is also  a divisor of $s$.
Since $x$ divides $m$, it also divides $n$, as $m \mid n$. Thus $x$ divides $t$ and $n$, and thus, by assumption, $x$ divides $s$ and $n$. In particular, it divides $s$, as requested. 
On the other hand, 2 does not appear to be true as stated. Take $t = 3, s = 2, n = 6, m = 2$.
Then
$$
3 = \gcd(3, 6) = \gcd(t, n) > \gcd(s, n) = \gcd(2, 6) = 2,
$$
but
$$
1 = \gcd(3, 2) = \gcd(t, m) < \gcd(s, m) = \gcd(2, 2) = 2.
$$

Even if you exchange $m$ and $n$ in 2, as suggested in a comment, it won't work. Take $t = 2, s = 3, n = 6, m = 2$.
Then
$$
2 = \gcd(2, 2) = \gcd(t, m) > \gcd(s, m) = \gcd(3, 2) = 1,
$$
but
$$
2 = \gcd(2, 6) = \gcd(t, n) < \gcd(s, n) = \gcd(3, 6) = 3.
$$
A: $\underbrace{(t,m)\mid m\mid\color{} n\ \&\ \color{}t}_{\Large\ \ \ (t,m)\,\ \mid\,\  \color{#c00}{n,\ t}} \,\Rightarrow\, \underbrace{(t,m) \mid (\color{#c00}{t,n})\mid\color{} s\ \& \ \color{} m}_{\Large\qquad\ (t,m)\,\ \mid\,\  \color{#0a0}{s,\ m}}\,$ $\Rightarrow$ $\, (t,m)\mid (\color{#0a0}{s,m})$
