SOLUTIONS

If M is the matrix (a,c;b,d) (rows (a,c) and (b,d)), then M maps the vector (1,0)^T to (a,b)^T and (0,1)^T to (c,d)^T. In order for the matrix to be in GL₂(R) we need in addition that it is nonsingular. So for A we get (0,c;1,d), with c not 0. Likewise B = (a,1;c,1) with a c, and C = (1,c;1,d) with c d.

1. Let g be an element in G. We show that R'_g is a bijection from G to G.

   R'_g is injective.
   Suppose R'_g(h) = R'_g(h'), then hg = h'g. By multiplying on the right-hand side by the inverse of g we find h = h'. Hence, R'_g is injective.

   R'_g is surjective.
   Let h be an element in G. Then R'_g(hg^-1) = h. Hence R'_g is surjective.

2. Only if G is commutative, for, if g and h do not commute, then R'_gR'_h(e) = hg is not equal to gh = R'_gh(e).

1. To show that f is a morphism of monoids we check that f(n + m) = f(n)f(m). For each k:
   f(n + m)(k) = k + 3(n + m) = (k + 3m) + 3m = (f(n)f(m))(k).

2. Since f(1)(1) = 4 we have 1 ~ 4, but clearly not 4 ~ 1.

If A is an orthogonal matrix, then the length of a column vector x and the product Ax are equal. So every orbit is contained in a set consisting of vectors of fixed length.

Let x be a column vector of length r > 0. Then there exists an orthogonal basis of Rⁿ whose first vector is x/r. The matrix formed by these column vectors is orthogonal and maps (r, 0, ..., 0) to x. So the vectors of length r are all in the orbit of (r, 0, ..., 0).

Of course, the orbit of the zero vector consists just of the zero vector.

• The stabilizer of the x-axis.
  A matrix A stabilizes the x-axis if the image of (1,0) is (a,0) for some nonzero a. Hence A is of the form (a,0;c,d) with ad nonzero.
• The stabilizer of the y-axis in H.
  In a similar way we find that the stabilizer (in H) of the y-axis consists of the matrices {(a,0;0,d) | ad nonzero}.
• The kernel of the action.
  The matrices fixing all lines also fix the x- and y-axis. So they are of the form (a,0;0,d). Since the line through (1,1) is also fixed, we get (a,0;0,a) with nonzero a. Such matrices indeed fix all lines through the origin.

H = <(1,2,3)> has two cosets in S₃, H and (1,2)H. The kernel is just H, the image S₂.

We consider the action of the automorphism group of the square on the vertices {1,2,3,4} in cyclic order. A basis is then the sequence [1,2]. The orbit of 1 is the full set {1,2,3,4}. The orbit of 2 in G₁ is {2,4}. So G has order 8. The action on the diagonals {13,24}: id, (13), (24) and (13)(24) fix the diagonals, the remaining permutations interchange them.

• (1,2,3,4)(1,2)(1,2,3,4)^-1 = (2,3)
• (1,2,3,4)(2,3)(1,2,3,4)^-1 = (3,4)
• (1,2,3,4)(3,4)(1,2,3,4)^-1 = (4,1)
• (1,2,3,4)(4,1)(1,2,3,4)^-1 = (1,2)

This implies that the group generated by (1,2,3,4) and a transposition distinct from (2,4) and (1,3) contains all transpositions and hence equals S₄. The group H is generated by two automorphisms of the square when points are labeled 1,2,3,4 in cyclic order, a reflection and a rotation over 90 degrees. This shows that H is isomorphic to the group D₄ of automorphisms of the quadrangle.

Label the vertices of the regular n-gon by 1,2, ... ,n. The n rotations (including the identity) are the n powers of the rotation (1, 2, ... ,n). The n reflections correspond to the maps r_k : i -> k-i (mod n), (k = 1, 2, ... ,n). A basis is given by the ordered pair [1,2], and the order of D_n is 2n.

• The tetrahedron.
  Let the points of the tetrahedron be 1,2,3,4. A basis for Aut(Tetrahedron) is [1,2,3].

  Orbit of 1: {1,2,3,4},
  Orbit of 2 in the stabilizer of 1: {2,3,4},
  Orbit of 3 in the stabilizer of 2: {3,4}.
  

  The order of the group is therefore 2 * 3 * 4 = 24.

• The cube.
  Let the points of the cube be 0 = {0,0,0}, 1 = {0,0,1}, 2 = {0,1,0}, etc, so i corresponds with the point whose coordinates are the binary digits (bits) of i. A basis for Aut(Cube) is [0,1,2].

  Orbit of 0: {0, ..., 7},
  Orbit of 1 in the stabilizer of 0: {1,2,4},
  Orbit of 2 in the stabilizer of 0,1: {2,4}.
  

  The order of the group is therefore 8 * 3 *2 = 48.

• Action on the diagonals.
  Every automorphism permutes the 4 diagonals, so produces a permutation on 4 symbols. The only nontrivial permutation that fixes the 4 diagonals is the `reflection in the center of the cube': It interchanges the end points of the diagonals. If two permutations induce the same action on the 4 diagonals, then they differ by this nontrivial permutation, or they are equal. This groups the 48 automorphisms in groups of 2. Therefore, we get all 24 permutations of the 4 diagonals.

Any automorphism of Q(a) has to fix 1 and thus Q, the subfield generated by 1.

The image of b of a under an automorphism has to satisfy the equation b⁴ - 2 = 0 (apply the automorphism to a⁴ - 2 = 0). That means b = xa where x = 1, -1, i or -i. Since Q(a) is contained in R, we only find x = 1, -1. This implies that there are at most 2 automorphisms of Q(a), namely the identity (clearly an automorphism) and the one determined by sending a to -a so that an arbitrary element f(a) is mapped to f(-a), where f is a polynomial (any element of the field can be expressed this way). It is easy to verify that this is an automorphism.

If gN = Ng for all g, then by multiplying both sides with the inverse of g, N = g^-1Ng. Hence g^-1ng is in N for all n in N and g in G. If g^-1Ng is contained in N then Ng is contained in gN for all g and hence g^-1N = g^-1(Ng) g^-1 is contained in g^-1(gN) g^-1 = Ng^-1 for all g. Hence Ng = gN for all g.

Let G be a non normal subgroup in H. (For example G = S₂ in H = S₂). Take for f the identity map.

1. K = {(1),(1,2)(3,4), (1,3)(4,2), (1,4)(2,3)}. Since K is a union of conjugacy classes it is normal in S₄.
2. Map (1,2)(3,4) to (1,0) and (1,3)(4,2) to (0,1) (for example).
3. {(1), (1,2), (3,4), (12)(3,4)}.

We recall L : G -> Sym(G/H), L_g(k H) = gkH.

1. If N is normal and contained in H and n is an element in N then L_n(k H) = nkH = k n₁H = k H.
2. |H| = 1, 3 or 9. If |H| = 1 or 9 then H is clearly normal. If |H| = 3 then |Sym(G/H)| = 6, so the kernel of the morphism L has order at least 3 and H = Ker(L) and therefore normal.

1. e forms a conjugacy class by itself
   
2. Two permutations in S_n are conjugate if and only if they have the same cycle structure.
3. If the group is commutative, then x y x^-1 = y, so that each conjugacy class consists of precisely one element.
   If for all x,y: x y x^-1 = y, then for all x, y: x y = y x so we have commutativity.
4. Follows from the identity (x y x^-1) ⁿ = x yⁿ x^-1.

1. Normality is immediate from the definition. Remains to show that it is a subgroup. If g,h are elements from Z, then xghx^-1 = xgx^-1xhx^-1 = gh and xg^-1x^-1 = (xgx^-1)^-1 = g^-1.
2. {(1)}.
3. The group itself.
4. The element itself.
5. xax^-1 is in <a> = {a,e}. But xax^-1 = e implies a = e, so xax^-1 = a for all x.
6. (aI)A = aA = A(aI) for all matrices A.
7. We prove the statement for GL_n(R). If A commutes with every nonsingular n by n matrix then A commutes with every n by n matrix (check this). Let E_ij be the matrix with zeros everywhere except in the ij position. Then AE_ij is a matrix with column i of A in column j and E_ijA has row j of A in row i. It follows that a_ii = a_jj, and all other entries are zero. Hence A is a scalar matrix.

1. An arbitrary element of Q/Z has the form a/b + Z for certain integers a and b. Adding the element to itself b times yields 0 + Z.
2. The elements 1/n + Z (for n = 1,2, ...) are all different.
3. 28/16 = 7/4 and the order is therefore 4.
4. The order of a/b is b/gcd(a,b).

Since Sym(4) does not contain elements of order 6 the same is true for its quotients. Hence Sym(4)/K is isomorphic to Sym(3).

1. This follows from |xy| = |x||y| and |x^-1| = |x|^-1.
2. It is clearly surjective. That it is a morphism follows from e^{a + b} = e^a· e^b
3. The kernel of the morphism f is Z. Application of the First Isomorphism Theorem does the rest.

1. The commutative groups are the cyclic groups and pruducts of cyclic groups.
2. D₄ contains two reflections whose product is a rotation over 90 degrees.
3. D₄ is in S₄. The other groups not.

Let G be a group of order 15. We show that G isomorphic to Z/15Z.

It suffices to show that there is an element of order 15. If this is not the case then all elements have order 1, 3 or 5. Moreover, G contains an element of order 3 and one of order 5. If a has order 5, then consider the action of G on the left cosets of H = <a>. Every element of order 5 has to be in the kernel of this action as there are only 3 cosets. Hence there is a unique (and hence normal) subgroup N of order 5 and the remaining 10 (non-identity) elements have order 3. The group G acts on N by conjugation. In this action an element of order 3 must commute with at least one of the 4 non identity elements of N. But the product of commuting elements of order 3 and 5 has order 15. Hence G is cyclic.

1. The left regular represention is defined by g -> L_g, with L_g(x) = gx. So we see that as permutations of {e,a,b,c}:
   L_e = id,
   L_a = (ea)(bc),
   L_b = (eb)(ac),
   L_c = (ec)(ab).
   
2. The action on X₂ of L_a is the permutation (eb,ac)(ec,ab) (the rest of the action follows by symmetry in a, b and c). The action cannot be transitive since |X₂| = 6 > 4 = |G|. In fact, there are three orbits, {ea,bc}, {eb,ac} and {ec,ab}.

1. The map C'_g is a bijection since it has an inverse, C'_g^-1.
   
2. This is only true in the special case that for all g, h, x the identity h^-1 g^-1xgh = g^-1 gh^-1xhg holds (for instance when G is commutative, or when G = D₄)
3. In this case we must have for all g, h, x the identity hgxhg = hgxgh. This is only true when G is commutative.

Two subsets of X are in the same orbit if and only if they have the same size. This is true both for S_n and A_n, except when n = 2, because then the two singletons {1} and {2} cannot be mapped to each other by A₂ (this is the trivial group).

1. We must show that (gh)x(i) = g(h(x))(i) for all g,h,x and i.
   (gh)x(i) = x((gh)^-1(i) = x(h^-1 g^-1(i)) = (h(x))(g^-1i) = (g(h))x(i).

2. There are two types of orbits. Vectors of the form (x,x) form one element orbits, other orbits have two elements and look like {(x,y), (y,x)}.
3. The stabilizer is S₂ for vectors of the form (x,x).
4. Geometrically (1,2) is a reflection in the line y = x.

1. We must show that (gh)x(i) = g(h(x))(i) for all g,h,x and i.
   (gh)x(i) = x((gh)^-1(i) = x(h^-1 g^-1(i)) = (h(x))(g^-1i) = (g(h))x(i).

2. Now there are three types of orbits, with 1, 3, 6 elements in them respectively. One element orbits consist of a vector having all coordinates equal; three element orbits consist of vectors having two coordinates the same; all other orbits contain the six vectors which can be obtained from (x,y,z) (three distinct coordinates) by applying the 6 permutations in S₃.
3. The stabilizer is S₃ for vectors (x,x,x).
   
4. Elements of order 2 correspond to certain reflections, elements of order three to rotations of 3-space, with axis the line x = y = z, and angle + or - 2/3.
5. The stabilizer is equal to (1,2) for vectors (x,x,y), with distinct x and y.

1. The only thing we have to show is that g(x) is again in R when x is. But this is obvious. The rest is a repetition of the argument from the previous two excercises.
2. We may associate to g the matrix M_g defined by M(i,j) = 1 if g(i) = j (and zero otherwise).
3. (1,2) is the reflection in the hyperplane x = y, (1,2,3,4) = (1,2)(2,3,4) and (2,3,4) as we saw in the previous exercise is a rotation with axis x₂ = x₃ = x₄.

1. We must show f((m,n) + (k,l)) = f(m + k,n + l). This follows from
   i^{(m + k) + (n + l)} = i^{m + n} · i^{k + l}.

2. The kernel is the set {(m,n) | m + n is divisible by 4}.
3. The orbit of z consists of the four points (if z not 0) {z, iz, -z, -iz}.
4. It is still obvious that it is a permutation representation. The kernel now is the set {(m,n) | m + n is divisible by 5}, the orbit of z (nonzero) consists of the five points {az | a⁵ = 1}. The orbit of 0 is {0}.

1. See exercise 26 and further.
2. Obvious.
3. The orbit of (1,1,0,0,0,0) consists of the 15 weight two vectors, so all vectors with two 1's and 4 zero coordinates.
4. Orbits are now the sets of vectors of fixed (even) weight. The number of vectors of weight 2k is n!/((2k)!(n - 2k)!), the total number of vectors of even weight is 2^{n - 1}.

1. The inverse of f(m) is easily checked to be f(-m), so every f(m) is a bijection. The map f is a morphism since f(m+n)(z) = f(m)f(n)(z) (= z i^{m + n}) for all m, n in Z and all z in S¹. The kernel of f is 4Z.
2. A square on S¹ has four points z, iz, -z, -iz, and they form precisely the orbit of z.
   
3. Elements of 2Z act as a reflection in the origin, or as the identity. The invariant sets are the pairs {z,-z}.

1. Since the map corresponding to A^-1 is the inverse of the one corresponding to A it is a bijection.
2. To see that it is injective we must show that {Au,Av} = {u,v} for all elements of X implies that A is the identity. But if A is not the identity, then fix a u with Au not equal to u nor equal to v, then {Au,Av} is not {u,v}.
3. 1. For every two bases of R³ we can find a linear transformation sending one to the other, so certainly for unordered triples of independent vectors.
   2. Same argument as before.

1. If we number the 3 cosets as follows: H= 3, (2,3)H = 2, and (1,3)H = 1, then L_(1,3) = (1,3) and L_(2,3) = (2,3). As S₃ is generated by (1,3) and (2,3), it follows that (with this numbering of the cosets), the map L is just the identity.
2. Do the same for the two groups H = < (1,2,3,4)> and < (1,2),(3,4)>. This is now a straightforward computation.

The orbit of an element x is the set {g(x) | g G}, and has obviously at most |G| = n elements.

Number the vertices of an octahedron as follows: 1 (top), 6 (bottom), and cyclically 2, 3, 4, 5 for the 4 vertices in the square in the middle. Then the following reflections are in the automorphism group G of the octahedron: a = (1,6), b = (3,5), c = (3,4)(2,5), d = (1,2)(4,6).

1. Use a, b, c, d to obtain that G is transitive.
2. The stabilizer of 1 (and then also 6) is the group of automorphisms of the square and has order 8. It is generated by b and c. The group therefore has order 6·8 = 48.
3. The group is transitive on the 3 diagonals as it is transitive on the vertices. The stabilizer of the diagonal {1,6} contains the element c which permutes the two other diagonals. Hence the whole S₃ is induced.
4. Yes, one can draw a cube through these 8 centers.
5. No, as there are adjacent and nonadjacent pairs.

Since G is transitive there is a g G with g(x) = x'. This g works: If h(x) = x, then

so that ghg^-1 is in G_x', etc.

1. a² - 2ia - 3 = (a - i - 2)(a - i + 2).
   So i = (a² - 3)/2a and is an element of Q(a).
2. 2 = a-i, so is also in Q(a).
3. Both inclusions are immediate from the above.
4. (a² - 2i a - 3)(a² + 2ia - 3) = a⁴ - 2a² + 9.
5. This polynomial has zeros +/-(i +/- 2).
6. The automorphism group is transitive on the roots that lie in the field, and (in this case) is determined by the image of a. So we get four automorphisms.
7. The group is Z/2Z × Z/2Z.

1. The roots are z, z², z³, z⁴ and z⁵ = 1.
2. Dividing by X - 1 yields the polynomial X⁴ + X³ + X² + X + 1.
3. The maps sending z to z^k, k = 1, 2, 3, 4 are the automorphisms. The automorphism group is isomorphic to Z/4Z.

1. Not normal.
2. Not normal, (13)(24) and (12)(34) are conjugate, but only the first one is in N.
3. Normal.
4. Normal.

The only nontrivial normal subgroup of S₃ is A₃

1. The diagonal matrices clearly form a subgroup. But if a is not b then (1,1;0,1)(a,0;0,b)(1,-1;0,1)=(a,-a + b;0,b) which is not a diagonal matrix.
2. In fact these matrices are in the center of G.
3. No, conjugate for instance with (0,1;1,0).

1. g^-1(N M)g = g^-1N g g^-1Mg = N M. (It is trivial that it is a subgroup).
2. It is again obvious that it is a subgroup. For h in H obviously h^-1N h = N, and h^-1H h = H, so the same goes for their intersection.
3. Almost anything works. Take H={(1),(12)(34)} for example. (Conjugate with (123)).
   
4. The group G = A₄ contains a normal subgroup H = < (1,2)(3,4),(1,3)(2,4)> of order 4. (Check that this subgroup is normal!) The group H is abelian, so every subgroup of it is normal. In particular, the group <(1,2)(3,4)> is normal in H but not in G.
5. It suffices to show that g^-1Hg is closed under taking products and inverses. g^-1hg g^-1kg = g^-1 hk g, and (g^-1hg) ^-1 = g^-1 h^-1 g.
6. Since g^-1 H g is a subgroup of the same cardinality as H it must be the same.

To prove that H is normal in G it suffice to prove that for all g in G and h in H the element g^-1hg is in H. Fix a g in G and an h in H. Since G is a finite group, every element from G, in particular g, is a finite product of elements from B. So, g can be written as

with the b_i in B.

Similarly, every element from H is a finite product of elements from H and h can be written as

with the a_i in B.

We will prove by induction on n that g^-1hg is in H. We first notice that

So, if n = 1, i.e., g an element of H, then the assumptions of the exercise imply that g^-1hg is a product of elements in H, and therefore itself is in H. This proves the first induction step.

Now suppose n > 1. Then

By the induction assumption, we can assume that (b_{n - 1}^-1 ··· b₁^-1 a_ib₁ ··· b_{n - 1}) is an element of H. Furthermore, now the first stap for n = 1 implies that b_n^-1 (b_{n - 1}^-1 ··· b₁^-1 a_ib₁ ··· b_{n - 1}) b_n is in H. This finishes the proof that H is normal in G.

1. First we show that f^-1(N) is indeed a subgroup. If a, b f^-1(N), then f(a) and f(b) and hence f(ab) = f(a)f(b) N. To see that it is normal, compute f(g^-1 a g) = f(g)^-1 f(a) f(g) and this is in N if a is.
2. Let h be an element in H then h = f(g) for some g in G. Hence h f(N) = f(g)f(N) = f(gN) = f(Ng) = f(N)f(g) = f(N)h. Hence f(N) is normal in H.

   The condition that F is surjective can not be removed. Indeed take for example G = N = S₂ and embed this via f naturally in H = S₃.

• Let g be an element of G not in H. Then the two cosets H and gH partition G. Similarly, the two cosets H and Hg partition G. Hence gH = Hg. This implies that H is normal in G.
• If we consider the permutation action of G on G/H, then we see that each element in H fixes one, and hence all (= 2) elements of G/H and therefore is in the kernel of the action. The elements outside H act nontrivially. Hence H is the kernel of the action and thus normal in G.
• If neither g nor h is in H, then gH = hH in the 2-element group G/H. But then the product gHhH = H, i.e., gh H.

1. G contains an element of order p. It generates a subsgroup H with index 2. So Exercise 45 applies.
2. Since 2 is a prime, G contains an element of order 2.
3. Trivial.
4. As soon as gh = hg for a single non identity element of H, then G is cyclic, and hence g commutes with all of them.
5. H is normal and g = g^-1 since its order is 2. The second statement follows from the commutativity of H.
6. gf = gghgh = hgh = hghgg = ghghg.
7. If some non-identity element h H commutes with g, then all of them do, since H is generated by each of its non-identity elements. If for some h then gh is not equal to hg, then ghgh = e, since it is in H and commutes with g.
8. Both G and D_p have the same order and are generated by elements g and h satisfying g² = h^p = e and ghgh = e.

Notice that for g in G and x in X we have gx^-1g^-1 = (gxg^-1)^-1. Hence without loss of generality we may assume that the set X is closed under taking inverses. Every element y in <X> is a product of elements from X. So y = x₁ ··· x_n. But then for every g in N_G(X) we have

which is again in <X>. Hence <X> is normal in N_G(X).

Let G be a normal subgroup of S₄.

• If G contains a transposition (i,j), then, since all transpositions are conjugate and generate S₄, we get the whole group.
• If G contains the product of two disjoint transpositions (i,j)(k,l), then it contains all such products. These together with e form a normal subgroup of order 4 isomorphic to Klein's fourgroup. If G contains in addition a 4-cycle, then it contains a transposition and we get the full group again. If G contains an additional 3-cycle then we get A₄, which is normal.
• If G contains any 3-cycle, it contains all, and again we get A₄ (or the whole group of course).
• Finally, if G contains a 4-cycle it also contains the product of two disjoint transpositions, and we get the whole group again.

1. Suppose h,k are elements in G_G(g). Then gh = hg and gk = kg. Hence g(hk) = (gh)k = (hg)k = h(gk) = h(kg) = (hk)g. So G_G(g) is closed under multiplication.

   Furthermore, multiplying gh = hg from the left and right with h^-1 yields h^-1g = gh^-1. So G_G(g) is also closed under taking inverses.

   Since the identity commutes with all elements, we find that G_G(g) is a subgroup of G.

2. If G is commutative, then G_G(g) = G.
3. If g does not commute with any element in G except the identity. This implies, for example, that conjugating g with any element except the identity never yields g again.
4. <(1,2)> × <(3,4)>.
5. Denote G_G(g) by C. Then h^-1gh = k^-1gk if and only if kh^-1g = gkh^-1 if and only if kh^-1 is in C if and only if k is in Ch. Hence the number of conjugates of g equals the number of right cosets of C in G.

We give a morphism such that the image is H and the kernel is N.

1. z -> |z|.
2. x -> x |x|.
3. z -> z |x|.
4. x -> (x mod m, x mod n).
5. 1 -> (0,0), i -> (1,0), j -> (0,1), k -> (1,1), -1 -> (0,0), -i -> (1,0), -j -> (0,1), -k -> (1,1).
6. A -> det(A)² is a morphism (use the properties of the determinant) to a cyclic group {1, 2, 4} of order 3 inside (Z/7Z)^*. This group of order 3 is of course isomorphic to Z/3Z.

1. This is obvious.
2. That S_n is n-transitive is also obvious. The n-2-transitivity follows easily from the fact that every sequence of n - 2 distinct elements can be extended to an even permutation.

1. It suffices to show that G contains all transpositions since these generate S_n. So let G contain the transposition (1,2) say, and consider another transposition (i,j). By 2-transitivity there is an element g G with g(1) = i and g(2) = j. But now g(1,2)g^-1 = (i,j) .
   
2. Essentially we do the same as before, but we need an intermediate step. So let G contain the 3-cycle (1,2,3), and we want to show that G contains an arbitrary 3-cycle (i,j,k). Conjugating with an element sending 1 to i and 2 to j we get the three cycle (i,j,l) for some l say different from k. We also can get a 3-cycle (l, k, m) for some m. If m is different from i and j we may conjugate (i,j,l) by (l,k,m) to get (i,j,k) and we are done. If by accident m = i, say, then conjugating (i,j,l) with the inverse, (l,i,k) does the job.

1. The kernel consists of the scalar matrices.
2. Since the action is obviously 2-transitive it suffices to find a matrix sending the 1 dimensional subspaces spanned by (1,0), (0,1), (1,1) to those spanned by (1,0), (0,1) and (a,b). This is achieved with the matrix (a,0;0,b).
3. If three 1-spaces are contained in a 2-space, the same is true for their images. Since for n > 2 not all triples of 1-spaces are contained in a 2-space the action cannot be 3-transitive.

Every single move is a 3-cycle and therefore even. So the moves generate a subgroup G of A₆. The order of A₆ is 6! / 2 = 360. Next we compute the order of our subgroup.

• The G-orbit of 1 is easily seen to have 6 elements, so | G | = 6 · | G₁ |, where G₁ is the stabilizer of 1.
• Next we consider the orbit of 2 under the stabilizer G₁ of 1. By using the three bottom triangles you easily see that 2, 3, 4, 5, 6 are in one G₁-orbit. So | G | = 6 · 5 · | G_1,2 |.
• Now turn to the orbit of 3 under the stabilizer G_1,2. Using the lower right triangle we see that 3, 5 and 6 are in one orbit. To see that 4 also belongs to this orbit, turn the lower left triangle once clockwise, and then turn the middle triangle once clockwise; the result is that 5 is moved to 4, showing that 4 is also in the orbit. Therefore, | G | = 6 · 5 · 4 · | G_1,2,3 |.
• The orbit of 4 under G_1,2,3 consists of 4, 5 and 6: 1) turn the lower left triangle clockwise, 2) turn the upper triangle counter clockwise, 3) turn the bottom right triangle clockwise, 4) turn the upper triangle clockwise. This produces the permutation (4,5,6), showing that 4, 5 and 6 are in one orbit.

We will show that the group G generated by these moves is the whole S₈. In particular, all positions can be obtained. We note the following.

• G is transitive on the vertices; this is easy.
• The stabilizer G₁ of 1 is transitive on the remaining 7 vertices. Again, this is easy.
• The stabilizer G_1,2 of 1 and 2 is transitive on the remaining 6 vertices: Use the top face and the back face.
• The stabilizer G_1,2,3 of 1, 2 and 3 is transitive on the remaining 5 vertices: turning the back face shows that 5, 6, 7, 8 are in one orbit; using the turns (1,4,8,5) (the left face), (5,6,7,8) (the back face) and then turning the left face back shows that 4 is also in that same orbit.
• The stabilizer G_1,2,3,4 of 1, 2, 3 and 4 is transitive on the remaining 4 vertices: Trivial!
• Conclusion: the group G is transitive on ordered 4-tuples.
• The group G contains the 4-cycle (1,2,3,4) and, as G is transitive on ordered 4-tuples, the conjugacy class of (1,2,3,4) in G consists of all 4-cycles.
• (1,2,3,4)(3,2,1,5) = (1,5,4) is in G. Hence G contains all 3-cycles and thus contains A₈.
• Since G contains an odd element, e.g., a 4-cycle, G is equal to S₈.