= Solution
The <group order> of a finite <group> is its number of elements, $|G|$. The <order of a group element> $g$ is the least positive <integer> $r$ with $g^r=e$. Such an $r$ exists in a finite group because two powers must coincide, and cancellation then gives a positive power equal to the <identity element>.
For the <direct product of groups>, the componentwise operation is closed because each factor is closed. <Associativity> holds in each coordinate:
$$
[(g,h)(g',h')](g'',h'')=(gg'g'',hh'h'')=(g,h)[(g',h')(g'',h'')].
$$
The <identity element> is $(e_1,e_2)$, and the <inverse element> of $(g,h)$ is $(g^{-1},h^{-1})$. Thus all <group axioms> hold. For finite factors, the <multiplication principle> gives
$$
\boxed{|G_1\times G_2|=|G_1||G_2|.}
$$
For $\mathbb Z_2\times\mathbb Z_2$, write $E=(e,e)$, $u=(w,e)$, $v=(e,w)$, $z=(w,w)$, with $w^2=e$. Its <Cayley table>, with the row element multiplying the column element, is
$$
\begin{array}{c|cccc}
&E&u&v&z\\\hline
E&E&u&v&z\\
u&u&E&z&v\\
v&v&z&E&u\\
z&z&v&u&E
\end{array}.
$$
This is the <Klein four-group>: every nonidentity element has order two.
In the <cyclic group> $\mathbb Z_4=\langle a\rangle$, the set $\{e,a^2\}$ is closed because $(a^2)^2=e$, and both elements have inverses in the set. It is therefore a proper <subgroup> of order two. By <Lagrange's theorem>, a proper subgroup has order one or two. An order-two subgroup must be generated by an element of order two. Neither $a$ nor $a^3$ has order two, whereas $a^2$ does. Hence
$$
\boxed{\text{the proper subgroups of }\mathbb Z_4\text{ are }\{e\}\text{ and }\{e,a^2\}.}
$$
\b[The PDF's claim needs “nontrivial”:] $\{e,a^2\}$ is the only nontrivial proper subgroup, but $\{e\}$ is also a proper subgroup under the usual definition.
A <group isomorphism> preserves element orders: $\phi(g)^r=e$ if and only if $g^r=e$, using the homomorphism property and injectivity. Since $\mathbb Z_4$ has elements of order four and the <Klein four-group> does not,
$$
\boxed{\mathbb Z_2\times\mathbb Z_2\not\cong\mathbb Z_4.}
$$
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