Group theory with direct product If $(G,*)$ and $(H,.)$ are groups, we can form a new group which is called the direct product $G \times H$ of $G$ and $H$, where the combination of two elements is defined by $(g_1,h_1)(g_2,h_2)=(g_1*g_2,h_1.h_2)$. Verify that $G \times H$ is a group. 
I know the group axioms are closure, associativity, inverse and identity. But this way of presenting a group is new to me, and I can't use the usual approach.
 A: You rely almost entirely on the fact that $G, H$ are both groups, and their operations are thus associative, closed, each has an identity, and each element of the respective group has an inverse.


*

*Closure Let $(g_1,h_1),(g_2,h_2)\in G\times H$ 
Then $(g_1,h_1)(g_2,h_2)=(g_1*g_2,h_1\cdot h_2)$ where $h_1\cdot h_2\in H$ and $g_1*g_2\in G$ so that, so $(g_1*g_2,h_1\cdot h_2) \in G\times H$

*Associativity:  Rely on the associativity of the respective operations of $G, H$. Let $g_1, g_2, g_3 \in G, h_1, h_2, h_3 \in H$ and show that $\Big((g_1,h_1)(g_2,h_2)\Big)(g_3,h_3)=(g_1, h_1)\Big((g_2, h_2)(g_3, h_3)\Big).$ Since the elements chosen are arbitrarily, this shows associativity holds over $G\times H$.

*Identity: Consider $(e_G,e_H) \in G\times H$, $e_G \in G$ the identity of $G$, $e_H \in H$ the identity in $H$. Show for all $(g, h) \in G\times H, (g, h)(e_G, e_H) = (g, h)$, etc.

*Inverses: Let  $(g,h)\in G\times H$ so we have that $(g^{-1},h^{-1})\in G\times H$ since both $H$ and $G$ are groups and so for all $g \in G, g^{-1} \in G$, and likewise for $h \in H$, $h^{-1} \in H$. Then $(g, h)(g^{-1}, h^{-1}) = (g * g^{-1}, h \cdot h^{-1}) = (e_G, e_H)$, etc. 
A: To prove associativity, just do the computation: 
$$\begin{align*}
\Big((g_1,h_1)(g_2,h_2)\Big)(g_3,h_3)&=(g_1*g_2,h_1\cdot h_2)(g_3,h_3)\\
&=\Big((g_1*g_2)*g_3,(h_1\cdot h_2)\cdot h_3\Big)\\
&=\Big(g_1*(g_2*g_3),h_1\cdot(h_2\cdot h_3)\Big)\\
&=(g_1,h_1)(g_2*g_3,h_2\cdot h_3)\\
&=(g_1,h_1)\Big((g_2,h_2)(g_3,h_3)\Big)\;.
\end{align*}$$
Each step is either from the definition of the group operation in $G\times H$ or from associativity of $*$ and $\cdot$ in their respective groups.
Everything else is an equally routine calculation using similar ideas.
A: So we can start by veryfiying that the identity is in the group by considering the element $(e_G,e_H)$......
We can show closure by taking $(g_1,h_1),(g_2,h_2)\in G\times H$ and then we have $(g_1,h_1)(g_2,h_2)=(g_1g_2,h_1h_2)$ where $h_1h_2\in H$ and $g_1g_2\in G$ so that.....
We can show we have inverses by taking $(g_1,h_1)\in G\times H$ and then we have that $(g_1^{-1},h_1^{-1})\in G\times H$ as both $H$ and $G$ are groups and so have inverses.....
