SOLUTIONS

For the permutations a = (1,2,3), b = (2,3,4,5,6), c = (1,4,6,3), we have

1. • a^-1 = (3,2,1),
   • abc = (1,2,3)(2,3,4,5,6)(1,4,6,3) = (1,5,6,4,3,2),
   • abc²= abcc = (1,5,6,4,3,2)(1,4,6,3) = (1,3,5,6,2),
   • c ^-1 b = (1,4,6,3 )^-1 (2,3,4,5,6) = (1,3,6,4)(2,3,4,5,6) = (2,6)(3,1)(4,5),
   • (acb) ^-1 = ((1,2,3)(1,4,6,3)(2,3,4,5,6))^-1 = ((3,6)(4,5,1)) ^-1 = (3,6)(1,5,4).

2. • sgn( a ^-1 ) = 1,
   • sgn(abc ) = - 1,
   • sgn(abc²) = 1,
   • sgn(c^-1b ) = -1,
   • sgn((acb) ^-1 ) = -1.

g(i) i implies g(g(i)) g(i), because g is injective. But then g(i) supp(g).

The permutation is of the form (a,b)(c,d,e), with a, ..., e all distinct. The 2-cycle can be chosen in 10 ways. Then, for the remaining 3-cycle, the three entries are determined, and so there are two distinct possibilities (there are six ways to order them, but each 3-cycle belongs to three such orderings). Hence there are 10 × 2 = 20 such elements.

1. As (a,b,c) = (a,b) (b,c) we have g = (1,2,3)(2,3,4)(3,4,5)(4,5,6)(5,6,7)(6,7,8)(7,8,9) = (1,2)(2,3) (2,3) ··· (7,8) (7,8)(8,9) = (1,2)(8,9).
2. The fixed points of g are those elements of {1,2, ..., 9} not occurring in a cycle of length >1 in the expression of g as a product of disjoint cycles. Thus, the fixed points are 3, 4, 5, 6, 7.
3. g^-1 = (9,8)(2,1) = (1,2)(8,9) (so g^-1 = g).
4. sgn(g) = (-1)² = 1, so g is even.

1. The inverse map of the map Z/nZ -> Z/nZ given by x -> ax is given by multiplication by a^-1 as is easily verified. So our map is a bijection.

   For a = 2 and n = 9 we find the disjoint cycle decomposition (1,2,4,8,7,5)(3,6).

2. If x is invertible mod n then ax is invertible.
   

   Likewise, if x is not invertible, then ax is not invertible. This means that if in a cycle of p_a the first element is invertible (resp. not invertible), the second, third, etc., element is also invertible (resp. not invertible).

Since a is an invertible element in R, its inverse a^-1 exists. First multiplying an element x of R by a and then by a^-1 (or vice versa) produces x again, because

Multiplication by a is therefore a permutation.

The elements of R are 0, 1, 2, a, a + 1, a + 2, 2a, 2a + 1, 2a + 2. Multiplying by a yields the following product of disjoint cycles (we have refrained from labelling the elements by the integers 1, 2, etc.):

Its cycle structure is 1,4,4.

1. If i supp(g), then g(i) supp(g). Since the supports of g and h are disjoint, both i and g(i) must then be fixed by h. Therefore:
   (gh)(i) = g(h(i)) = g(i), and (hg)(i) = h(g(i)) = g(i).

   Similarly, if i supp(h), then

   (gh)(i) = g(h(i)) = h(i), and (hg)(i) = h(g(i)) = h(i).

   If i belongs to neither supp(g), nor supp(h), then g(i) = h(i) = i. Consequently,

   (gh)(i) = i, (hg)(i) = i.

2. In the product (gh)^m exactly m factors g and m factors h occur. Using the previous item repeatedly, we move all the factors g to the left. This leaves us with g^mh^m. (Formalists may want to prove the statement by induction.)
3. As in the first item, we derive that g^m(h^m(i)) = g^m(i) if i is in the support of g, that g^m(h^m(i)) = h^m(i) if i is in the support of h, and that g^m h^m(i) = i if i is in the support of neither (i.e., is fixed by both). Combining this with g^mh^m = id, we find g^m(i) = i for all i and similarly h^m(i) = i for all i. This means that g^m = h^m = id.
4. Apply division with remainder to m and t to obtain m = qt + r, with 0 r < t. It follows that
   g^m = (g^t)q · g^r = g^r,

   so that g^r = id. Since r < t and t is the order of g, this only leaves the possibility r = 0. But then t divides m.

5. Suppose
   g = c₁ c₂ ··· c_q

   is a product of disjoint cycles and has cycle structure m₁, m₂, ..., m_r. Let m = lcm(m₁, m₂, ..., m_r). We first show that g^m = id. The second item implies

   g^m = c₁^m c₂^m ··· c_q^m.

   If c_i is an l-cycle, then l occurs among the m_j, so that m is a multiple of l. But then

   c_i^m = (c_i^l)^m/l = id.

   In conclusion, g^m = id.

   Next we show that g^s = id implies that s is a multiple of m. If g^s = id then we find

   c₁^s c₂^s ··· c_q^s = id

   and we conclude c₁^s = id , ..., c_q^s = id. This implies that s is divisible by m₁, m₂, ..., m_r, and therefore by lcm(m₁, m₂, ..., m_r).

   In conclusion, the smallest positive number such that g^m = id is lcm(m₁, m₂, ..., m_r).

1. Notice that (1,2) = (1,2,3,4) (1,3,4,2) (1,2,3,4). Since we can conjugate (1,2) to each transposition we want, we can write each transposition as a product of 4-cycles. As each permutation is a product of transpositions, it is also a product of 4-cycles.

2. Notice that (3,5,4) = (1,2,3,4,5) (2,1,5,3,4). As above we now find that each even permutation can be written as a product of 5-cycles.

Take for b the permutation (1,9,4,6,2,8)(3,7,5).

Let g S_n with n > 2.

1. Suppose that g commutes with the transposition (i,j), where i j. Then
   g(i,j)g^-1 = (g(i),g(j)) = (i,j).

   Therefore, g(i) {i, j}.

2. By the first part, if g commutes with both (i,j) and (i,k), then g(i) is a member of {i, j} {i, k} and thus equals i.

3. An element g that commutes with all elements of S_n commutes with all different transpositions and so, by the previous part, will fix every i.

A₄ is the set of even permutations of S₄. The identity is even. The cycle structure of any other even permutation in S₄ must be 2,2 or 3. There are 3 permutations of cycle structure 2,2 and 8 of cycle structure 3 in A₄. Thus, we find
A₄ = {e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3), (1,2,3), (1,2,4), (1,3,2), (1,3,4), (1,4,2), (1,4,3), (2,3,4), (2,4,3)}.

Let a=(1,2) and b=(2,3, ..., n).

1. bab^-1 = (1,3).
2. b^kab^-k = (1,k +1) for k < n.
3. For i j we have (1,i)(1,j)(1,i) = (i, j). Each element of S_n may be written as a product of transpositions. The above shows that each transposition and then also each permutation can be written as a product using only the elements a, b and b^-1.

1. We leave this to the reader. Do you see a relation with switching columns or rows?
2. First note that every row and every column of a permutation matrix contains exactly one entry 1, the other entries are 0. Let K be the permutation matrix of gh. We compute K_i,j for arbitrary i and j and compare with (MN)_i,j. There are unique numbers k and m such that i = g(k) and k = h(m). Now K_i,j = 1 if i = (gh)(j) and 0 otherwise. In the first case j = m. Now (MN)_i,m is the sum over the products
   M_i,tN_t,m.

   Such a product is only nonzero if i = g(t) and t = h(m), and in that case it equals 1. In the second case, j m and all products M_i,tN_t,j are 0, so that (MN)_i,j = 0. Consequently, K = MN.

3. If g is the transposition (k,l), then the permutation matrix M is obtained from the identity matrix by switching the k-th and l-th column. But then det(M) = -det(I) = -1, where I is the identity matrix.

4. Suppose g is a permutation with permutation matrix M. Then g can be written as a product of transpositions

   g = t₁ ··· t_k.

   The permutation matrix M is now the product of the permutation matrices M_i of the t_i.

   M = M₁ ··· M_k.

   But since both sgn and det are multiplicative, we find that sgn(g) = det(M).

Number the vertices of a quadrangle clockwise, starting at the top left.

1. The rotation through +90° whose center is the middle point of the quadrangle is described by the permutation (1,2,3,4). The reflection in the diagonal through the vertices 1 and 3 is described by the permutation (2,4).
2. The permutations mapping an edge onto an edge are
   • (1)(2)(3)(4)    the identity permutation,
   • (1,2)(3,4)    reflection with axe through the middle of 12 and the center of the quadrangle,
   • (1,4)(2,3)    reflection with axe through the middle of 14 and the center of the quadrangle,
   • (1,3)(2,4)    rotation through 180 whose center is the center of the quadrangle,
   • (1,2,3,4)    rotation through +90 whose center is the center of the quadrangle,
   • (2,4)    reflection in the diagonal through the vertices 1 and 3,
   • (1,3)    reflection in the diagonal through the vertices 2 and 4,
   • (1,4,3,2)    rotation through -90 whose center is the center of the quadrangle,
3. The first 4 of the above permutations are even, the others odd. In the previous part it is already described which are the reflections and which the rotations.

Suppose A is an n by n matrix with entries A_i,j. Then the entry A^T_i,j equals A_j,i.

The determinant of A is the sum over all g S_n of the products

and det(A^T) is the sum over all g S_n of the products

since A^T_i,j = A_j,i. Consider the product

Each pair of indices (i,g(i)) can be written as the pair (g^-1(k),k), where k = g(i). As g is a bijection, this implies that each of the n factors in the product

occurs as a factor of

and conversely. But then

because sgn(g) = sgn(g^-1). The last step is to remark that if g runs through S_n, g^-1 also runs through S_n. Putting everything together yields

det(A) = _g sgn(g) A_1,g(1) A_2,g(2) ··· A_n,g(n) = _g sgn(h) A_g(1),1 A_g(2),2 ··· A_g(n),n = det(A^T),

where the sum is over all g S_n.

1. You easily check that you get the 4 permutations e, (1,2)(3,4), (1,3)(2,4) and (1,4)(2,3). If you also allow a reflection in the main diagonal, you get 4 more permutations: (1,3,4,2), (1,2,4,3), (2,3) and (1,4).
2. A few obvious moves are (1,2), (2,4), (3,4). In particular, we find (1,2) and (3,4)(2,4) = (2,3,4). Now use Exercise 12 to conclude that you can produce every permutation in S₄.
3. Consider the three rows and columns. It is easily seen that each move moves rows to rows and columns to columns, although the order of the entries in a row or column is not necessarily preserved. The same then holds for compositions of moves. Each composition of moves is therefore determined by what it does with the three rows (a permutation of 1, 2, 3) and what it does with the three columns (again a permutation of 1, 2, 3). This produces a bijection of our set of permutations to the set S₃ × S₃.
4. Each individual move consists of a 3-cycle, so has even sign. As a consequence, the permutations you can get this way are contained in A₉. In fact, you can produce every even permutation using these moves, but that is much easier to explain with the theory of Chapter 8.

1. (1,2,3), (3,2,1), (1,2,4), (4,2,1), (2,3,4), (4,3,2), (1,3,4), (4,3,1).
2. Consider an arbitrary vertex j with new label g(j) and let i be the vertex such that h(i) = j. Since h moves i to j, it will move the new label g(i) of vertex i to vertex j. The resulting permutation therefore maps j to g(i). As i = h^-1(j), we find that j is mapped to gh^-1(j).
3. Applying a move means multiplication on the right-hand side by the inverse of the permutation corresponding to the move. Each move is a 3-cycle, so has even sign, just like its inverse. Therefore the result of each sequence of moves is an even permutation. So we get at most the 12 even permutations in A₄. To see that all 12 arise, list them as follows:
   • there is the identity e;
   • for every face there are two nontrivial moves, producing 8 permutations;
   • for each pair of faces (there are 3 of them) a suitable combination of the two moves corresponding to these faces yields (1,2)(3,4), (1,3)(2,4) and (1,4)(2,3).
   (Using the results of Chapter 6 it is much easier to reach this conclusion.)