I have to show that for all $(h_1, g_1) \in H_1 \times G_1 $ and all $(h, g) \in H \times G$ it is the case that $(h,g)(h_1, g_1)(h,g)^{-1} \in H_1 \times G_1$. We know that $(h, g)^{-1} = (h^{-1}, g^{-1})$ and so we have $(h, g)(h_1, g_1)(h, g)^{-1} = (hh_1h^{-1}, gg_1g^{-1})$ and from assumption we have $hh_1h^{-1}\in H_1, gg_1g^{-1} \in G_1$. Is it enough to prove this statement?
Morover I would like to show that $(H \times G)/(H_1 \times G_1) \cong (H/H_1) \times (G/G_1)$. I think it can be shown from fundamental theorem on homomorhpisms by showing $f : H \times G \rightarrow (H/H_1) \times (G/G_1)$ such that $ker f = H_1 \times G_1$ and $im f = (H/H_1) \times (G/G_1)$, but I don't know how to define such $f$.
Please help me with these questions. Thank you.
Define $f : H \times G \rightarrow (H/H_1) \times (G/G_1)$ as $$f(h,g)=(hH_1,gG_1),\forall (h,g)\in H\times G.$$ That is $f=(\pi_H,\pi_G)$, where $\pi_H:H\to\frac{H}{H_1}$ and $\pi_G:G\to\frac{G}{G_1}$ are quotient homomorphisms. Hence, $f$ is also a homomorphism.
Now, $$\begin{align} (h,g)\in\text{ker}(f)&\iff(hH_1,gG_1)=(e_HH,e_GG)\\ &\iff (h,g)\in H_1\times G_1,\\ \end{align}$$
which implies $\text{ker}(f)=H_1\times G_1.$
Clearly, $f$ is surjective. So by first fundamental theorem of groups we have, $$\frac{H\times G}{H_1\times G_1}=\frac{H\times G}{\text{ker}(f)}\simeq\text{Im}(f)=(H/H_1) \times (G/G_1).$$