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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}

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