theorem Th54:
  A |^ a |^ a" = A & A |^ a" |^ a = A
proof
  thus A |^ a |^ a" = A |^ (a * a") by Th47
    .= A |^ 1_G by GROUP_1:def 5
    .= A by Th52;
  thus A |^ a" |^ a = A |^ (a" * a) by Th47
    .= A |^ 1_G by GROUP_1:def 5
    .= A by Th52;
end;
