This can be shown directly; for fun, though, we will prove it using
Cayley's isomorphism ϕ.
First note that for
f,g bijective, we have
(f∘g)−1=g−1∘f−1
So given
a,b∈G this gives
(ϕ(a)∘ϕ(b))−1ϕ(ab)−1ϕ((ab)−1)ϕ((ab)−1)(ab)−1=ϕ(a)−1∘ϕ(b)−1=ϕ(a)−1∘ϕ(b)−1=ϕ(a−1)∘ϕ(b−1)=ϕ(a−1b−1)=a−1b−1
Note that step
(2→3) follows from the property of
group isomorphisms that
ϕ(a)−1=ϕ(a−1).