Part 1.
Part 2.
Part 3.
Part 4.
(ab)N = {abx | x 
N} =
a{bx | x N} =
a(bN).
N} =
a{bx | x N} =
a(bN).
aNN =
{axy | x, y N} =
{az | z N} =
aN.
N} =
{az | z N} =
aN.
As N is a normal subgroup, Nb = bN, so by Steps 1 and 2,
aNbN = a(Nb)N = a(bN)N =
(ab)NN =
(ab)N = abN.
Suppose a bN.
Then aN is a subset of bN. But, as these cosets are classes
of an equivalence relation (see a previous lemma), they coincide.
bN.
Then aN is a subset of bN. But, as these cosets are classes
of an equivalence relation (see a previous lemma), they coincide.