If $G=\langle g \rangle$ and $H=\langle h \rangle$, show that $G\times H=\langle(g,1),(1,h)\rangle$.
I don't really know what it is I am supposed to do. My guess is that I let $G$ be defined as above then somehow show that $G=\{(g_1,1),(g_2,1)\dots\}$, but I don't know how to do so.
Note that, by definition, $$ G\times H = \{(g,h): g\in G, h\in H\}. $$ This is a group under the operation $$(g,h)(g',h') = (gg', hh').$$ Clearly, then, $$ \langle(g,1), (h,1)\rangle \subseteq G\times H . $$ You just need to prove that $\langle(g,1), (h,1)\rangle \supseteq G\times H$. Now let $(g,h)\in G\times H$. Well, then $(g,h) = (g,1)(1,h)$, so ...