If you mean: does knowing the *G*_{i} tell you the Galois group of *P*, then no.

Examples:

$P = (X^2+1)(X^2-2)$ has Galois group $C_2 \times C_2$, and both factors have Galois group $C_2$; this works because the splitting fields of the two factors intersect only in $\mathbb{Q}$.

But $P = (X^2 + X + 1)(X^2+3)$ has Galois group $C_2$, although both factors again have Galois group $C_2$. Here both factors, though they're coprime, define the same extension $\mathbb{Q}(\sqrt{-3})$.

I've just seen Robin's answer, so to relate to that: in the first example, the Galois group of *P* is the whole of $G_1 \times G_2$. In the second example, it is the diagonal subgroup of $G_1 \times G_2$, which is smaller although still projects surjectively onto each factor.