theorem Th36:
  [.a,b.] = 1_G iff a * b = b * a
proof
  thus [.a,b.] = 1_G implies a * b = b * a
  proof
    assume [.a,b.] = 1_G;
    then (b * a)" * (a * b) = 1_G by Th17;
    then a * b = (b * a)"" by GROUP_1:12;
    hence thesis;
  end;
  assume a * b = b * a;
  then
A1: (b * a)" = b" * a" by GROUP_1:18;
  [.a,b.] = (b * a)" * (a * b) by Th17;
  hence [.a,b.] = (b" * a") * a * b by A1,GROUP_1:def 3
    .= b" * b by GROUP_3:1
    .= 1_G by GROUP_1:def 5;
end;
