theorem Th19:
  [.1_G,a.] = 1_G & [.a,1_G.] = 1_G
proof
  thus [.1_G,a.] = ((1_G)" * a") * a by GROUP_1:def 4
    .= (1_G * a") * a by GROUP_1:8
    .= a" * a by GROUP_1:def 4
    .= 1_G by GROUP_1:def 5;
  thus [.a,1_G.] = (a" * (1_G)") * a by GROUP_1:def 4
    .= (a" * 1_G) * a by GROUP_1:8
    .= a" * a by GROUP_1:def 4
    .= 1_G by GROUP_1:def 5;
end;
