Let $H_1$, $H_2$, $H_3$ be normal subgroups of a group $G$. Furthermore let
- $H^1=\langle H_2\cup H_3\rangle$,
- $H^2=\langle H_1\cup H_3\rangle$,
- $H^3=\langle H_1\cup H_2\rangle$.
Suppose that the homomorphism $p:G\rightarrow G/H^1\times G/H^2\times G/H^3$ defined by $$p(g):=(gH^1,gH^2,gH^3)$$ is an isomorphism of $G$ onto $G/H^1\times G/H^2\times G/H^3$.
Do these assumptions imply that $G$ is the internal direct product of $H_1$, $H_2$ and $H_3$?
Yes:
Notice $H_1\le H^2\cap H^3$ and as $\rho$ is an isomorphism $H^1\cap H^2\cap H^3=1$ so $H_1\cap H^1=1$. Therefore $G/H^1$ has a subgroup isomorphic to $H_1$. Similarly $G/H^2$ has a subgroup isomorphic to $H_2$ and $G/H^3$ has a subgroup isomorphic to $H_3$.
Together this means $G$ has a subgroup isomorphic to $H_1\times H_2\times H_3$ which is the image of $H_1H_2H_3$ under $\rho$.
If $g\in G\setminus H_1H_2H_3$ then suppose $\rho(h)=(gH_1,H_2,H_3)$ for some $h\in G$. In particular $h\in H_2\cap H_3$ and $h^{-1}g\in H_1$ so $g=h(h^{-1}g)\in H_1H_2H_3$ contrary to assumption. Hence $\rho$ is not surjective contrary to assumption. This means $G=H_1H_2H_3$ which is isomorphic under $\rho$ to $H_1\times H_2\times H_3$.