theorem Th23:
  (a * b) |^ g = a |^ g * (b |^ g)
proof
  thus (a * b) |^ g = g" * (a * 1_G * b) * g by GROUP_1:def 4
    .= g" * (a * (g * g") * b) * g by GROUP_1:def 5
    .= g" * (a * g * g" * b) * g by GROUP_1:def 3
    .= g" * (a * g * (g" * b)) * g by GROUP_1:def 3
    .= g" * (a * g) * (g" * b) * g by GROUP_1:def 3
    .= a |^ g * (g" * b) * g by GROUP_1:def 3
    .= a |^ g * (b |^ g) by GROUP_1:def 3;
end;
