Here is the problem I'm working on.
Let $G,H$ be groups. Let $\Gamma = \{(g,e_{H}) | \ g \in G\}$ and $\Sigma = \{(e_{G},h) | \ h \in H\}$. Show that for any $x = (g,h) \in G \times H$ we have $x\Gamma x^{-1} = \Gamma$ and $x\Sigma x^{-1} = \Sigma$.
I don't think I even understand what the question is saying: what does it mean to say that $x\Gamma x^{-1}$? Does it mean applying the group operation of $G$ to any elements $x, (g, e_{H}),x^{-1}$ like this: $(g_0,h_0)(g,e_H)(g_0^{-1},h_0^{-1})$ and therefore considering their triple direct-product?
As an example, the question gives the hint that the Chinese remainder theorem shows that $C_n \times C_m$ is isomorphic to $C_{nm}$ if $gcd(n,m) = 1$ but I don't find this particularly informative.