theorem
  for H being strict Subgroup of G holds H * a = G implies H = G
proof
  let H be strict Subgroup of G;
  assume
A1: H * a = G;
  now
    let g;
    assume
A2: not g in H;
    now
      assume g * a in H * a;
      then ex h st g * a = h * a & h in H by Th58;
      hence contradiction by A2,ThB16;
    end;
    hence contradiction by A1;
  end;
  hence thesis by A1,Th62;
end;
