reserve n for Element of NAT;
reserve i for Integer;
reserve G,H,I for Group;
reserve A,B for Subgroup of G;
reserve N for normal Subgroup of G;
reserve a,a1,a2,a3,b,b1 for Element of G;
reserve c,d for Element of H;
reserve f for Function of the carrier of G, the carrier of H;
reserve x,y,y1,y2,z for set;
reserve A1,A2 for Subset of G;

theorem Th2:
  for X being Subgroup of A, x being Element of A st x = a holds
  x * X = a * (X qua Subgroup of G) &
  X * x = (X qua Subgroup of G) * a
proof
  let X be Subgroup of A, x be Element of A;
  set I = X qua Subgroup of G;
  assume
A1: x = a;
  thus x * X c= a * I
  proof
    let y be object;
    assume y in x * X;
    then consider g being Element of A such that
A2: y = x * g & g in X by GROUP_2:103;
    reconsider h = g as Element of G by GROUP_2:42;
    a * h = x * g by A1,GROUP_2:43;
    hence thesis by A2,GROUP_2:103;
  end;
  thus a * I c= x * X
  proof
    let y be object;
    assume y in a * I;
    then consider b such that
A3: y = a * b and
A4: b in X by GROUP_2:103;
    reconsider c = b as Element of X by A4;
    reconsider c as Element of A by GROUP_2:42;
    a * b = x * c by A1,GROUP_2:43;
    hence thesis by A3,A4,GROUP_2:103;
  end;
  thus X * x c= I * a
  proof
    let y be object;
    assume y in X * x;
    then consider g being Element of A such that
A5: y = g * x & g in X by GROUP_2:104;
    reconsider h = g as Element of G by GROUP_2:42;
    h * a = g * x by A1,GROUP_2:43;
    hence thesis by A5,GROUP_2:104;
  end;
  thus I * a c= X * x
  proof
    let y be object;
    assume y in I * a;
    then consider b such that
A6: y = b * a and
A7: b in X by GROUP_2:104;
    reconsider c = b as Element of X by A7;
    reconsider c as Element of A by GROUP_2:42;
    b * a = c * x by A1,GROUP_2:43;
    hence thesis by A6,A7,GROUP_2:104;
  end;
end;
