theorem
  g.(a |^ i) = (g.a) |^ i
proof
  per cases;
  suppose
A1: i >= 0;
    hence g.(a |^ i) = g.(a |^ |.i.|) by ABSVALUE:def 1
      .= (g.a) |^ |.i.| by Th36
      .= (g.a) |^ i by A1,ABSVALUE:def 1;
  end;
  suppose
A2: i < 0;
    hence g.(a |^ i) = g.(a |^ |.i.|)" by GROUP_1:30
      .= (g.(a |^ |.i.|))" by Th32
      .= ((g.a) |^ |.i.|)" by Th36
      .= (g.a) |^ i by A2,GROUP_1:30;
  end;
end;
