We can represent \(c\) with prime factorization:
$$ c = {p_1}^{z_1} * {p_2}^{z_2} * {p_3}^{z_3} * {p_4}^{z_4} * {p_5}^{z_5}* \ldots * {p_n}^{z_n} $$
\(p_i\) is the \(i\)th prime integer and \(p_n\) is the largest prime integer before \(\max \{ a,b,c \} \). It's also possible for \(z_i\) to be 0. Since \(\gcd (a, b) = 1\), then the prime factorizations of \(a\) and \(b\) won't have any common prime integers. Let's give an example of the prime factorization of \(a\), \(b\) and \(ab\):
$$ \begin{align} a &= {p_1}^{x_1} * {p_4}^{x_4} * {p_6}^{x_6}* \ldots * {p_{n-1}}^{x_{n-1}} \\ b &= {p_2}^{y_2} * {p_3}^{y_3} * {p_5}^{y_5} * \ldots * {p_n}^{y_n} \\ ab &= {p_1}^{x_1} * {p_2}^{y_2} * {p_3}^{y_3} * {p_4}^{x_4} * {p_5}^{y_5} * {p_6}^{x_6}* \ldots *{p_n}^{y_n} \end{align} $$
If we take the \(\gcd \) with \(c\):
$$ \begin{align} \gcd(a, c) &= {p_1}^{\min \{x_1,z_1\} } {p_4}^{\min \{x_4,z_4\} } {p_6}^{\min \{x_6,z_6\} }\ldots {p_{n-1}}^{\min \{x_{n-1},z_{n-1}\} } \\ \gcd(b, c) &= {p_2}^{\min \{y_2,z_2\} } {p_3}^{\min \{y_3,z_3\} } {p_5}^{\min \{y_5,z_5\} } \ldots {p_n}^{\min \{y_n,z_n\} } \\ \gcd(ab, c) &= {p_1}^{\min \{x_1,z_1\} } {p_2}^{\min \{y_2,z_2\} } {p_3}^{\min \{y_3,z_3\} } {p_4}^{\min \{x_4,z_4\} } {p_5}^{\min \{y_5,z_5\} } {p_6}^{\min \{x_6,z_6\} }\ldots {p_n}^{\min \{y_n,z_n\} } \end{align} $$
Looking at the prime factorizations above, it is trivial that \(\gcd(ab, c) = \gcd(a, c) * \gcd(b, c)\).