= Solution
{c}
Choose distinct $a,b\ne e$, possible because $|G|>2$. By <6/a/iii>, $a$ and $b$ commute, and $a^2=b^2=e$. The four elements
$$
H=\{e,a,b,ab\}
$$
are distinct: $ab=e$ would give $a=b$, while $ab=a$ or $ab=b$ would make $b=e$ or $a=e$. Commutativity and $a^2=b^2=e$ show that products stay in $H$, and every element is its own inverse. More explicitly,
$$
\boxed{C_2\times C_2\longrightarrow H,\qquad(i,j)\longmapsto a^ib^j\quad(i,j\in\{0,1\})}
$$
is a <group isomorphism>, with addition of exponents modulo two. Hence $G$ contains a <subgroup> isomorphic to the <Klein four-group>.
The converse is false. The <direct product of groups> $C_2\times C_2\times C_3$ contains $C_2\times C_2\times\{e\}$, but it has elements of <order of a group element> three. Thus the existence of the specified <subgroup> does not force every nonidentity element to have <order of a group element> two.
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