= Solution
Interpret the functions as maps on $\mathbb R\setminus\{0\}$, so every reciprocal is defined. Let $e(x)=x$, $a(x)=-x$, $b(x)=1/x$ and $c(x)=-1/x$. With the left factor applied last, direct <function composition> gives the <Cayley table>
$$
\boxed{\begin{array}{c|cccc}
\circ&e&a&b&c\\\hline
e&e&a&b&c\\
a&a&e&c&b\\
b&b&c&e&a\\
c&c&b&a&e
\end{array}.}
$$
The table proves closure and identifies $e$ as the identity. Every element is its own inverse. Associativity is inherited from composition of maps, so the set is a <group>. The three nonidentity elements are commuting <involutions>.
Its subgroups are
$$
\boxed{\{e\},\quad\{e,a\},\quad\{e,b\},\quad\{e,c\},\quad\{e,a,b,c\}.}
$$
There are no others: a subgroup containing two distinct nonidentity elements also contains their product, the third, and hence is the whole group; a subgroup containing just one has order two.
The abstract <dihedral group> $D_2$ has commuting generators $R,m$ with $R^2=m^2=I$, for example a rectangle's half-turn and reflection in one axis. The assignment $I\mapsto e$, $R\mapsto a$, $m\mapsto b$, $Rm\mapsto c$ respects products and is bijective. Thus
$$
\boxed{F\cong D_2\cong C_2\times C_2.}
$$
This is the <group generated by sign reversal and reciprocation>, a realization of the <Klein four-group>. The nonzero domain is essential; treating these as functions defined at zero would not give the stated group.
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