We will show that ~ is an equivalence relation and leave the proof of the second statement to the reader. Thus, we need to establish that ~ is

reflexive,

symmetric,

transitive.

g ~ g, since g^-1g = e H.

If g ~ k, then k^-1g H. But then (k^-1g)^-1 = g^-1k is also in H and we find k ~ g.

If g ~ k and k ~ h, then k^-1g, h^-1k H. But then also (h^-1k)(k^-1g) = h^-1g H. Thus g ~ h.