Problem source

Let $(G,*)$ and $(H,\circ)$ be two groups and $G\times H$ be the
set of ordered pairs $(g,h)$ with $g\in G$ and $h\in H.$ A multiplication
on $G\times H$ is defined by 
\[
(g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2})
\]
for all $g_{1},g_{2}\in G$ and $h_{1},h_{2}\in H$. 
Show that, with this multiplication, $G\times H$ is a group. 
State whether the following are true or false and prove your answers. 
\begin{questionparts}
\item $G\times H$ is abelian if and only if both $G$ and $H$ are abelian. 
\item $G\times H$ contains a subgroup isomorphic to $G$. 
\item $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ is isomorphic to $\mathbb{Z}_{4}.$
\item $S_{2}\times S_{3}$ is isomorphic to $S_{6}.$
\end{questionparts}
{[}$\mathbb{Z}_{n}$ is the cyclic group of order $n$, and $S_{n}$
is the permutation group on $n$ objects.{]} 

Solution source

Claim: $G \times H$ is a group. (Called the product group).

Proof: Checking the group axioms:
\begin{enumerate}
\item (Closure) is inherited from $G$ and $H$, since $g_1 * g_2 \in G$ and $h_1 \circ h_2 \in H$
\item (Associativity) 
\begin{align*}
(g_1, h_1)\l (g_2, h_2)(g_3,h_3)\r &= (g_1, h_1)(g_2 *g_3, h_2 \circ h_3) \\
&= (g_1*(g_2 *g_3), h_1 \circ (h_2 \circ h_3)) \\
&= ((g_1*g_2) *g_3), (h_1 \circ h_2) \circ h_3) \\
&= (g_1*g_2, h_1 \circ h_2)(g_3, h_3) \\
&= \l(g_1, h_1)(g_2, h_2) \r(g_3,h_3)
\end{align*}
\item (Identity) Consider $(e_G, e_H)$, then $(e_G, e_H)(g,h) = (g,h) = (g,h)(e_G, e_H)$
\item (Inverses) If $(g,h) \in G \times H$ then consider $(g^{-1}, h^{-1})$ and we have $(g^{-1}, h^{-1})(g,h) = (e_G,e_H) = (g,h)(g^{-1}, h^{-1})$
\end{enumerate}

\begin{itemize}
\item Claim: $G \times H$ is abelian iff $G$ and $H$ are. 
Proof: $\Rightarrow$ Suppose $g_1, g_2 \in G$ and $h_1, h_2 \in H$ then $(g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (h_1,h_2)(g_1, g_2) = (g_2 * g_1, h_2 \circ h_1)$ so $g_1*g_2 = g_2*g_1$ and $h_1 \circ h_2 = h_2 \circ h_1$, therefore $G$ and $H$ are commutative.
$\Leftarrow$ If $H$ and $G$ are commutative then:
$(g_1, g_2)(h_1,h_2) = (g_1 * g_2, h_1 \circ h_2) = (g_2 * g_1, h_2 \circ h_1) = (h_1,h_2)(g_1, g_2)$ so $G \times H$ is commutative.

\item Claim: $G\times H$ contains a subgroup isomorphic to $G$. Consider the subset $S = \{(g,e_H) : g \in G \}$. Then this is a subgroup isomorphic to $G$ with isomorphism given by $\phi  : S \to G$ by $\phi((g,e_H)) = g$

\item If $x \in \mathbb{Z}_2 \times \mathbb{Z}_2$ then $x^2 = e$, but $1$ does not have order 2 in $\mathbb{Z}_4$

\item $S_2 \times S_3$ has order $2 \times 6 = 12$. $S_6$ has order $6! \neq 12$

\item 
\end{itemize}