First I suuuuuck at proofs. I think I am on the right track but I need some fine tuning. Or if I am totally off let me know. First we show that $H_i$ is nonempty. Note that since $H_i$ is a subgroup of $G_i, H_i$ contains the identity element. So $e_G \in H_1, e_G \in […]

- Tags ..., ...x_n \in G_n, ...x_n \in H_n.$ Since $H_i \leq G_i, 'Y_2', $ which shows that $(x_1, e_G \in H_2, e_G \in H_n.$ This means that $e_G \in H_1 \times H_2 \times \cdots \times H_n.$ Hence $H_i$ is non empty. Next we show it is a subset. Let $, First I suuuuuck at proofs. I think I am on the right track but I need some fine tuning. Or if I am totally off let me know. First we show t, H_i$ contains the identity element. So $e_G \in H_1, x_1 \in G_1, x_2, x_2 \in G_2, x_2 \in H_2, x_2y_2^{-1}, x_2y_2^{-1} \in H_2, x_n) \in H_1 \times H_2 \times...\times H_n.$ This means $x_1 \in H_1, x_n) \in H_1 \times H_2 \times...\times H_n$ and therefore $\in G_1 \times G_2 \times...\times G_n.$ Hence $ H_1 \times H_2 \times...\times H, x_n)(y_1^{-1}, x_n)$ and $y= (y_1, x_ny_n^{-1} \in H_n.$ This means that $(x_1y_1^{-1}, x_ny_n^{-1}) \in H_1 \times H_2 \times \cdots \times H_n$ since $H_i \leq G_i$ and inverses are elements in the group. Hence $H_1 \times H_2, x_ny_n^{-1}).$ Then $x_1y_1^{-1} \in H_1, y_2^{-1}, y_n^{-1})= (x_1y_1^{-1}, y_n) \in H_1 \times H_2 \times...\times H_n.$ Then $xy^{-1}= (x_1