Subgroup generated by two subsets of $G^{\omega_1}$ Let $H$ be a subgroup of the group $G$ and $M$ be the $\Sigma$-product of $\omega_1$ copies of the group $G$. Consider the isomorphism $i$ of $G$ to the diagonal $\Delta$ of $G^{\omega_1}$, now put $H^*=i(H)$.

If $E$ be the smallest subgroup of $G^{\omega_1}$ containing both $M$ and $H^*$; How can we construct an element $h\in E$ by elements of $M$ and $H^*$? 

It follows from the definition of $E$ that $H^*\subset \Delta\cap E$. But is it true that $\Delta\cap E\subset H^*$ ?

$\newcommand{supp}{\operatorname{supp}}$For each $x=\langle x_\xi:\xi\in\omega_1\rangle\in M$ let $\supp(x)=\{\xi\in\omega_1:x_\xi\ne 1_G\}$. Suppose that $x\in M$ and $h\in H$. Then $i(h)x=\langle hx_\xi:\xi\in\omega_1\}$; if we let $i(h)x=\langle y_\xi:\xi\in\omega_1\rangle$, then $y_\xi=h$ for all $\xi\in\omega_1\setminus\supp(x)$, while the $y_\xi$ with $\xi\in\supp(x)$ can be anything in $G$. The situation with $xi(h)$ is similar. Let 
$$E_0=\left\{x\in G^{\omega_1}:\exists h\in H\Big(\left|\{\xi\in\omega_1:x_\xi\ne h\}\right|\le\omega\Big)\right\}\;;$$
clearly $E_0\subseteq E$, and I claim that $E=E_0$. It suffices to show that $E$ is closed under multiplication and taking inverses, and those are both clear.
In other words, an element of $G^{\omega_1}$ belongs to $E$ iff it differs from an element of $H^*$ on at most countably many coordinates. It follows that $\Delta\cap E=H^*$.
