To show that the relation itso is indeed an equivalence relation we have to check that the relation is
reflexive
symmetric
transitive
For every x 
X,
we have ex = x,
so x itso x.
X,
we have ex = x,
so x itso x.
Suppose x itso y. Then there is g
G with
g(x) = y.
Consequently,
x = g-1(y).
As g-1
G,
this implies that
y itso x.
G with
g(x) = y.
Consequently,
x = g-1(y).
As g-1
G,
this implies that
y itso x.
Suppose x itso y and y itso z. Then
there are g, h G such
that g(x) = y and h(y) = z.
Now hg(x) = h(g(x)) =
h(y) = z, and hg
G, so x itso z.
G such
that g(x) = y and h(y) = z.
Now hg(x) = h(g(x)) =
h(y) = z, and hg
G, so x itso z.