Said in more detail, we have a group $G$, and $H_1,\cdots,H_n$ are its subgroups such that $$G\cong H_1\times\cdots H_n:=\Gamma$$ Then I want to prove that $G$ is exactly (not just isomorphic to) the internal direct product of them, that's to say $$G=H_1H_2\cdots H_n$$ with $H_1,\cdots, H_n$ being normal subgroups of $G$ and $$H_i\cap H_1\cdots H_{i-1}H_{i+1}\cdots H_n=\{\text{id}_G\}$$
For me the most difficult part is I don't know how to figure out what the isomorphism from $\Gamma$ to $G$ is, although I sense it can't be very arbitrary since $H_i$s are subgroups of $G$. If I denote by $\phi:\Gamma\to G$ the isomorphism from $\Gamma$ to $G$, and by $\tilde{H_i}$ the embedment of $H_i$ into $\Gamma$, then $\Gamma=\tilde{H_1}\cdots \tilde{H_n}$. Clearly this is an internal direct product, and I can therefore show that $G$ is the internal direct product of $\phi(\tilde{H_i})$s, but how can I show that $\phi(\tilde{H_i})=H_i$?
The simplest counterexample is for $n=2$ with $G=C_2\times C_2$, with $H_1=H_2=C_2\times\{1\}$.
But maybe you want $H_1\cap H_2=\{1\}$ and $G=H_1H_2$, in which case another example is $G=S_3\times C_2$ with $H_1=\{(\sigma,x)\}\cong S_3$ where $x=1$ iff $\sigma\in A_3$, and $H_2$ a cyclic subgroup of order $2$ of $S_3\times\{1\}$. Since $H_2$ is not normal in $G$, $G$ is not the internal direct product of $H_1$ and $H_2$.