Is it always true that if $\gcd(a,b)=1$ then $\gcd(ab, c) = \gcd(a, c)\gcd(b, c)$? We had a discussion in lecture recently about the greatest common denominator, and the topic came up about, given $\gcd(a, b) = 1$ and any nonzero integer $c$, whether or not $$\gcd(ab, c) = \gcd(a, c)\gcd(b, c).$$ Is this always true? We also talked about the general case $\gcd(ab, c) = \gcd(a, c)\gcd(b, c)$, and how that is not always true, but this other problem has interested me. Can anyone offer some insight?
 A: $\: (a,c)(b,c) = (a(b,c),c(b,c)) = (ab,ac,bc,cc) = (ab,(a,b,c)c),  $ which $= (ab,c)$ if $\:(a,b,c) = 1$. Thus it is certainly true in your case $\:(a,b)=1,\:$ but not generally, e.g. it fails for  $\:a = b = c > 1$.
Remark $\ $  We can go further to obtain a precise condition for equality:
Theorem $\quad (a,c)(b,c) = (ab,c) \iff (a,b,c,ab/(ab,c)) = 1$
Proof $\quad $ Continuing from the above calculation  we have
$\begin{eqnarray}\quad  (a,c)(b,c) 
&=&\qquad (c\ (a,b,c),ab)\ &&\ \ \text{by above} \\
&=&\, ((ab,c)\,(a,b,c),ab) &&\ \ \text {by a known gcd law [1] }\\
&=&\ \ (ab,c)((a,b,c),ab/(ab,c)) && \ \ \text{by factoring out}\ (ab,c)
\end{eqnarray}$
and the above $ =\, (ab,c)$ iff the 2nd factor $= 1.\ $ QED $ $ $ $  Here's said link [1]
A: The answer is yes.
Let 
$$\gcd(ab,c)= p_1^{\alpha_1}  \cdots p_n^{\alpha_n}$$
Then for each $i$ we have
$$p_i^{\alpha_i}\mid c \,;\, p_i^{\alpha_i}\mid ab$$
Since $a,b$ are relatively prime, we get $ p_i^{\alpha_i}\mid a$ or $ p_i^{\alpha_i}\mid b$. Thus $p_i^{\alpha_i\mid \gcd(a,c)}$ or $p_i^{\alpha_i}\mid\gcd(b,c)$. 
This proves that 
$$\gcd(ab,c) \mid \gcd(a,c) \gcd(b,c)$$
For the other implication, the key is to observe that 
$$\gcd( \gcd(a,c), \gcd(b,c)) \mid \gcd(a,b)=1$$
Since $\gcd(a,c)\mid\gcd(ab,c)$, $ \gcd(b,c)\mid\gcd(ab,c)$ and the two numbers are relatively prime, their product divides $\gcd(ab,c)$.
A: Consider $$a = 2 \\ b = 2 \\ c = 2 $$
A: Let $d_1=gcd(ab,c)$, $d_2=gcd(a,c)$ and $d_3=gcd(b,c)$. For any prime $p$, let $ord_p m$ denote the highest power of $p$ that divides $m$. Want to show that for any prime $p$, $ord_p d_1 = ord_p (d_2 d_3)$. It is trivial to see that $ord_p (mn)=ord_p m + ord_p n$. This means WTS $ord_p d_1 = ord_p d_2 +ord_p d_3$. 
Suppose there exists $p$ prime s.t.$ord_p d_1 > ord_p d_2 +ord_p d_3$ . Then if $ord_p d_1=k$ then $p^k|ab$ and $p^k|c$. Since $gcd(a,b)=1$, the first tells us that either $p^k|a$ or $p^k|b$. In the former case $p^k$ is a common divisor for $a$ and $c$ and in the latter for $b$ and $c$. This means in the former case $p^k|d_2$ implying $ord_p d_2\geq k$ clearly a contradiction, since $ord_p$ is a non-negative function and LHS>RHS. It also means, in the latter case $p^k|d_3$ implying $ord_p d_3\geq k$ a contradiction for the same reason. Suppose there exists $p$ prime s.t.$ord_p d_1 > ord_p d_2 +ord_p d_3$ . It is easy to argue that if $ord_p d_2 \neq 0$ then $ord_p d_3=0$, and the other way around. If not, you would have that for some prime $p$, $p$ divides both $d_1$ and $d_2$ and therefore, both $a$ and $b$ a contradiction. WLOG suppose $ord_p d_2=k\neq 0$. We have that $p^k|d_2$ implying $p^k|a$ and $p^k|c$. This means $p^k|ab$ and therefore $p^k|d_1$. This implies $ord_p d_1\geq k = ord_p d_2+ord_p d_3$. This is a contradiction with the fact that $ord_p d_1 > ord_p d_2 +ord_p d_3$. It follows therefore that for all primes $p$ $ord_p d_1 = ord_p (d_2 d_3)$. Therefore the equality is true. 
