 Solution

Choose distinct $a,b\ne e$, possible because $|G|>2$. By iii, $a$ and $b$ commute, and $a^2=b^2=e$. The four elementsare 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,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.