Proof
We need to show that, if, for all g, h 
G,
we have
f(gh) = f(g)f(h), then we also have
f(e) = e'.
f(x-1) = f(x)-1
for all x G.
f(x)f(x-1) =
f(xx-1) =
f(e) = e' by the previous part.
Similarly,
f(x-1)f(x) =
e'.
Thus, f(x-1) is the inverse in G' of
f(x).