= Solution
Write $G=\{e,a,b,c\}$. By <Lagrange's theorem>, the <order of a group element> divides four. Since no element has order four, all three nonidentity elements have order two. Thus every element is its own inverse. For any $x,y\in G$,
$$
xy=(xy)^{-1}=y^{-1}x^{-1}=yx,
$$
so the <group> is <Abelian>. Also $ab$ cannot equal $e,a$ or $b$, by cancellation and the fact that $a,b$ are distinct nonidentity elements. Hence $ab=c$. The remaining products are forced, giving
$$
\boxed{\begin{array}{c|cccc}
&e&a&b&c\\\hline
e&e&a&b&c\\
a&a&e&c&b\\
b&b&c&e&a\\
c&c&b&a&e
\end{array}}
$$
The map $e\mapsto(0,0)$, $a\mapsto(1,0)$, $b\mapsto(0,1)$, $c\mapsto(1,1)$ preserves this table under componentwise addition modulo two. Therefore
$$
\boxed{G\cong C_2\times C_2=V,}
$$
the <Klein four-group>, also called the Vierergruppe.
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