= Solution
By <Lagrange's theorem>, the possible values of the <order of a group element> are
$$
\boxed{1,2,4,8}.
$$
These are possibilities across groups of order eight; they need not all occur in one group.
If every element has order at most two, every element equals its own inverse. Thus for any $x,y$,
$$
xy=(xy)^{-1}=y^{-1}x^{-1}=yx,
$$
so $G$ is an <abelian group>. Choose $a\ne e$, and then $b\notin\langle a\rangle$. The four elements $e,a,b,ab$ are distinct and form a <subgroup> $H$: commutativity and $a^2=b^2=e$ prove closure and inverses. Choose $c\notin H$. The elements of $H$ and $Hc$ are disjoint, since $hc\in H$ would imply $c\in H$. These two four-element sets therefore exhaust $G$.
Every element consequently has a unique expression $a^ib^jc^k$, with $i,j,k\in\{0,1\}$. The map from triples to these products is bijective and respects multiplication by addition of exponents modulo two. Hence
$$
\boxed{G\cong C_2\times C_2\times C_2},
$$
a <direct product of groups> and an <elementary abelian group>.
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