Brian Bi
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Section 6.7. Abstract Symmetry: Group Operations

Exercise 6.7.1

1. $$D_4$$ consists of the identity, three non-identity rotations, two reflections across diagonals, and two reflections across lines parallel to two sides. By inspection, the stabilizer of a vertex consists only of the identity and the single reflection across the diagonal that contains that point, and the stabilizer of an edge consists only of the identity and the single reflection across a line perpendicular to that edge.
2. By inspection, the stabilizer consists of the identity operation, the 180 degree rotation, and the reflections across the two diagonals.

Exercise 6.7.2 Pick some point $$P$$ on $$\ell$$ arbitrarily. Every plane isometry can then be uniquely written in the form $$m = t_a \rho_\theta r^k$$ where $$k$$ is 0 or 1, the rotation is considered to be around the point $$P$$, and the reflection is considered to be across the line $$\ell$$. Since $$r^k$$ itself stabilizes $$\ell$$ for either choice of $$k$$, the orientation-preserving part $$t_a\rho_\theta$$ must also stabilize $$\ell$$ for $$m$$ to stabilize $$\ell$$ overall. In the orientation-preserving part, the rotation must be through an angle of either 0 or $$\pi$$, otherwise it would change the line's slope. Since $$P$$ lies on $$\ell$$, either $$\rho_0$$ or $$\rho_\pi$$ itself stabilizes $$\ell$$, so the translation part must stabilize $$\ell$$ also. We therefore conclude that the stabilizer of $$\ell$$ is $$\{t_a \rho_{i\pi} r^k \mid a \parallel \ell, i \in \{0, 1\}, k \in \{0, 1\}\}$$.

Exercise 6.7.3

1. The action of $$S_3$$ on $$U$$ is a homomorphism $$S_3 \to S_3$$ since $$|U| = 3$$. The kernel of this homomorphism can only be the trivial group or the normal subgroup $$A_3$$. In the latter case the image would have order 2 so it would have to consist of the identity and a transposition, and the element not involved in the transposition would be a fixed point. We are told that the action on $$U$$ is transitive, so this latter possibility is ruled out. The action of $$S_3$$ on $$U$$ is therefore an automorphism $$S_3 \to S_3$$ that maps the three transpositions $$(1\ 2), (2\ 3), (3\ 1)$$ to three different transpositions of $$U$$. Label the elements of $$U$$ fixed by these three transpositions, respectively, as $$u_3, u_2, u_1$$. Then each transposition $$\tau \in S_3$$ acts by sending $$u_i$$ to $$u_{\tau(i)}$$ for each $$i$$, and since the transpositions generate $$S_3$$, the same is true for general permutations $$\sigma \in S_3$$. We can use the action of $$S_3$$ on $$V$$ to label the elements of $$V$$ in the same manner. The diagonal action of $$\sigma$$ on $$U \times V$$ therefore maps $$(u_i, v_j)$$ to $$(u_{\sigma(i)}, v_{\sigma(j)})$$. If $$i = j$$, the orbit therefore consists of the three elements $$(u_1, v_1), (u_2, v_2), (u_3, v_3)$$. On the other hand, if $$i \neq j$$ and $$i' \neq j'$$, then we can always find a permutation $$\sigma$$ with $$\sigma(i) = i', \sigma(j) = j'$$, which would therefore map $$(u_i, v_j)$$ to $$(u_{i'}, v_{j'})$$. This implies that the other 6 elements of $$U \times V$$, that is, the elements of the form $$(u_i, v_j)$$ with $$i \neq j$$, form a single orbit.
2. Label the elements of $$U$$ as in part (a). The action of $$S_3$$ on $$V$$ is a homomorphism $$S_3 \to G$$ where $$G$$ is a subgroup of $$S_3$$ that fixes the index 1 and does not fix 2 and 3; $$G$$ can only be $$\{1, \tau_{23}\}$$. Since $$G$$ has order 2, the kernel of this homomorphism is $$A_3$$, and therefore even permutations of $$S_3$$ act as the identity on $$V$$ while odd permutations interchange $$v_2$$ and $$v_3$$. Obviously the orbit of $$(u_i, v_1)$$ under the diagonal action can only contain elements of the form $$(u_j, v_1)$$, and since $$S_3$$ acts transitively on $$U$$, one of the orbits is $$U \times \{v_1\}$$. On the other hand, suppose $$i, i' \in \{1, 2, 3\}$$ and $$j, j' \in \{2, 3\}$$. Then we can always find $$\sigma \in S_3$$ with $$\sigma(u_i, v_j) = (u_{i'}, v_{j'})$$, since there are two permutations sending $$i$$ to $$i'$$ of different parities and we choose the even or odd one depending on whether $$j$$ and $$j'$$ are equal or unequal. Therefore the other 6 elements $$\{(u, v) \mid u \in U, v \in V \setminus \{v_1\}\}$$ form a single orbit.

Exercise 6.7.5 Assume the cube is positioned so that each of its faces is parallel to one of the coordinate planes.

There are 24 orientation-preserving symmetries of the cube, which can be described as follows:

• the identity operation,
• 4 rotations through each of the angles $$2\pi/3$$ and $$4\pi/3$$, taken around body diagonals, for a total of 8 operations,
• 3 rotations through each of the angles $$\pi/2, \pi, 3\pi/2$$, taken around the coordinate axes, for a total of 9 operations,
• 6 rotations through the angle $$\pi$$, taken around the lines joining midpoints of opposite edges, for a total of 6 operations.

We know that these $$1 + 9 + 8 + 6 = 24$$ symmetries are distinct since two rotations around different axes are different rotations, as are two rotations around the same axis with different angles. So we have described all 24 orientation-preserving symmetries.

The 24 orientation-reversing symmetries can be classified as:

• reflections across the (3) coordinate planes and across the (6) planes that contain corresponding face diagonals on opposite faces, for a total of 9 operations,
• the inversion operation,
• 8 rotoreflections consisting of a reflection across a plane perpendicular to a body diagonal followed by a rotation through angle $$\pi/6$$ or $$5\pi/6$$ around that body diagonal, and 6 rotoreflections consisting of a reflection across a coordinate plane followed by a rotation through angle $$\pi/2$$ or $$3\pi/2$$ around the perpendicular axis, for a total of 14 operations.

We argue that these $$9 + 1 + 14 = 24$$ orientation-reversing symmetries are distinct (and therefore account for all the orientation-reversing symmetries) as follows. A reflection fixes all points on the plane of reflection, while the inversion and rotoreflection operations have only a single fixed point at the origin. An inversion has the entirety of $$\mathbb{R}^3$$ as an eigenspace, while a rotoreflection where the angle of rotation is neither 0 nor $$\pi$$ has only a single eigenvector, which is parallel to the axis. So the three categories in the list above are disjoint. Two reflections in different planes are distinct, so there are indeed 9 reflections among the symmetries of the cube. Two rotoreflections with different axes are also distinct since they have different eigenvectors, as are two rotoreflections with the same axis and different angles. So there are indeed 14 rotoreflections among the symmetries of the cube.

Exercise 6.7.6 Assume the prism is placed so that its centroid lies at the origin and the triangular faces are parallel to the xy plane.

Consider first the orientation-preserving symmetries, which consist of the identity operation and pure rotations. A rotation that is a symmetry of the prism must stabilize the z-axis, since the latter is determined by the centroids of the two triangular faces. The rotation must therefore have $$\hat{z}$$ as an eigenvector. There are two cases:

1. The eigenvalue of $$\hat{z}$$ is $$+1$$. This implies that the rotation is around the z-axis itself. If it is not the identity, then it must be by 120 or 240 degrees.
2. The eigenvalue of $$\hat{z}$$ is $$-1$$. This implies that:
• the angle of rotation is 180 degrees;
• the axis of rotation is perpendicular to the z-axis;
• the two rays that constitute the z-axis are exchanged by the rotation, and therefore so are the two triangular faces.
In order for such a rotation to map a vertex of the prism to another vertex (which, as stated, will be on the opposite triangular face), the axis of rotation must contain the midpoint of the two vertices. By considering the three possible target vertices of a single vertex under such a rotation, we see that the axis must pass through the centroid of one of the rectangular faces; and a rotation of 180 degrees around either of these three axes is a symmetry of the prism. So there are 3 rotational symmetries in this category.

This makes a total of 6 orientation-preserving symmetries of the triangular prism. The prism also has at least one plane of symmetry, such as the plane parallel to and equidistant from the two triangular faces; let $$r$$ denote the reflection across this plane. Then the map $$g \mapsto rg$$ is a bijection between orientation-preserving symmetries to orientation-reversing symmetries, so there are also exactly 6 orientation-reversing symmetries, for a total of 12. A rectangular face is stabilized by 4 of these symmetries: the identity, the rotation around an axis that passes through its centroid, and the composition of $$r$$ with these two symmetries.

Exercise 6.7.7

1. Obviously one orbit consists of the zero vector alone. We claim that all other vectors lie in a single orbit. To see this, first observe that all $$n$$ standard basis vectors lie in a single orbit (they are mapped to each other by permutation matrices). Now let $$v \in \mathbb{R}^n \setminus \{0\}$$. Pick $$i$$ such that $$v_i \neq 0$$. Then $$v = Av'$$ where $$v' = 1$$ and $$A$$ is an elementary matrix of the third kind that scales entry $$i$$ by the factor $$v_i$$. And $$v'$$ can be written as a product of elementary matrices of the first kind and $$e_i$$ where each elementary matrix of the first kind is used to add a multiple of element $$i$$ to one of the other elements. Since all elementary matrices are invertible, this establishes that every nonzero vector lies in the orbit of some standard basis vector, and since all standard basis vectors lie in a single orbit as well, this implies that all nonzero vectors form a single orbit.
2. $$Me_1 = e_1$$ if and only if the first column of $$M$$ is $$e_1$$. The stabilizer of $$e_1$$ therefore consists of all invertible matrices that have $$e_1$$ as their first column. Using expansion by minors, we can characterize this set as the set of $$n \times n$$ matrices with $$e_1$$ as their first column, arbitrary elements in the other entries in the first row, and an invertible matrix for the remaining $$(n-1) \times (n - 1)$$ entries.

Exercise 6.7.8

1. If $$A, B$$ are invertible, then they lie in the same orbit, since $$A = (AB^{-1})B$$ and $$AB^{-1} \in GL_2(\mathbb{C})$$. It is also clear that an invertible matrix is not in the same orbit as any singular matrix. So one orbit consists of precisely the invertible matrices. The zero matrix also forms its own orbit. The remaining matrices in $$\mathbb{C}^{2\times 2}$$ have rank 1 and we claim that the matrices $A_x = \begin{pmatrix} 1 & x \\ 0 & 0 \end{pmatrix}, \qquad B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},$ all belong to different orbits and that these orbits exhaust the matrices of rank 1. To see this, observe that $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} ae & af \\ ce & cf \end{pmatrix}$ so left-multiplication won't change the ratio of the entries in the first row. Furthermore an arbitrary rank 1 matrix can be row-reduced to the form $$A_x$$ or $$B$$ as follows. If the first column of $$M$$ vanishes, the second column must not vanish; the rows can be swapped if needed to make the top-right entry nonzero, then the first row can be scaled to make the top-right entry 1, and a multiple of the first row can be added to the second row to zero out the lower-right entry, resulting in matrix $$B$$. Performing elementary row operations is equivalent to left-multiplying by an invertible matrix, so $$M$$ lies in the same orbit as $$B$$. If on the other hand the first column of $$M$$ doesn't vanish, then we can swap the rows as need to make the top-left entry nonzero, then scale the first row to make the top-left entry equal to 1, then add a multiple of the first row to the second row to zero out the bottom-left entry. These elementary row operations don't change the rank of the matrix, so the resulting matrix must have 0 in its bottom-right entry in order to still have rank 1. Therefore the resulting matrix is some $$A_x$$, and $$M$$ and $$A_x$$ lie in the same orbit.
2. According to the Jordan normal form theorem, there is one orbit for each Jordan form up to permutation of the blocks. Therefore, there is a separate orbit for every pair of distinct complex numbers $$\lambda_1, \lambda_2$$, with $$\diag(\lambda_1, \lambda_2)$$ as a representative, and for every complex number $$\lambda$$ there are additionally two distinct orbits: one with $$\lambda I$$ as a representative, and one whose representative is $$\lambda I$$ plus a 1 in the upper-right corner.

Exercise 6.7.9

1. The identity element of $$G$$ is $$(I, I)$$, and $$(I, I) * A = IAI^{-1} = A$$. Also, $$((P_1, Q_1)(P_2, Q_2))A = (P_1 P_2)A(Q_1 Q_2)^{-1} = P_1 P_2 A Q_2^{-1} Q_1^{-1} = P_1(P_2 A Q_2^{-1}) Q_1^{-1} = (P_1, Q_1) * ((P_2, Q_2) * A)$$. So $$*$$ is a valid group action.
2. The action of $$G$$ allows us to multiply by any invertible matrix on either the left or right, therefore the orbit of a matrix consists of those matrices that can be reached by any sequence of elementary row operations and elementary column operations, in any order; left-multiplication performs row operations while right-multiplication performs column operations. Each matrix of rank $$r$$ can be converted to reduced row echelon form with $$r$$ leading coefficients and then column operations can be used to move each leading coefficient onto the diagonal and clear all remaining entries. Thus every $$m \times n$$ matrix of rank $$r$$ lies in the same orbit as the matrix that has $$1$$ for the first $$r$$ entries of the diagonal and 0 at all other entries. This, together with the fact that row and column operations preserve rank, implies that there is one orbit of $$*$$ for each possible rank $$r = 0, 1, \ldots, \min(m, n)$$.
3. Write $$Q$$ in block form $Q = \left[\begin{array}{c|c} A & B \\ \hline C & D \end{array}\right]$ where $$A$$ is an $$m \times m$$ matrix, $$B$$ is an $$m \times (n - m)$$ matrix, $$C$$ is an $$m \times (n - m)$$ matrix, and $$D$$ is an $$(n - m) \times (n - m)$$ matrix. We want to find $$P, Q$$ such that $P \left[\begin{array}{c|c} I & 0\end{array}\right] \left[\begin{array}{c|c} A & B \\ \hline C & D \end{array}\right]^{-1} = \left[\begin{array}{c|c} I & 0\end{array}\right]$ Multiply both sides on the right by $$Q$$ to obtain $\left[\begin{array}{c|c} P & 0\end{array}\right] = \left[\begin{array}{c|c} I & 0\end{array}\right] \left[\begin{array}{c|c} A & B \\ \hline C & D \end{array}\right]$ therefore $$P = A$$ and $$B = 0$$. The stabilizer therefore consists of pairs of an invertible $$m \times m$$ matrix $$P$$ and an $$n \times n$$ matrix of the form $Q = \left[\begin{array}{c|c} P & 0 \\ \hline C & D \end{array}\right]$ where $$C$$ is an arbitrary $$(n - m) \times m$$ matrix and, in order for $$Q$$ to be invertible, $$D$$ is an invertible but otherwise arbitrary $$(n - m) \times (n - m)$$ matrix.

Exercise 6.7.10

1. According to the Jordan normal form theorem, the orbit consists of those matrices that have the given matrix as their Jordan normal form, namely, those whose eigenvalues are 1 and 2. As for the stabilizer, consider conjugation by some matrix $$M$$ with determinant $$\Delta \neq 0$$, \begin{align*} M\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} M^{-1} &= \frac{1}{\Delta} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \\ &= \frac{1}{\Delta} \begin{pmatrix} ad - 2bc & ab \\ -cd & 2ad-bc \end{pmatrix} \\ &= \begin{pmatrix} 1 - \frac{bc}{\Delta} & \frac{ab}{\Delta} \\ -\frac{cd}{\Delta} & 1 + \frac{ad}{\Delta} \end{pmatrix} \end{align*} so in order to get the original matrix back we must have $$bc = ab = cd = 0, ad = \Delta$$. That is, the stabilizer consists of the invertible diagonal matrices.
2. This is fairly straightforward to do using the counting formula. The order of $$GL_2(\mathbb{F}_5)$$ is 480: the first column can be one of 24 nonzero vectors, and each choice excludes 5 possible choices for the vector in the second column, namely those that are multiples of the first column, leaving 20. According to part (a), the stabilizer consists of the 16 invertible diagonal matrices. (Note that the argument for the form of the stabilizer did not depend on the Jordan normal form theorem, which only holds in algebraically closed fields.) So the order of the orbit is $$480 \div 16 = 30$$. I'm not sure how Artin expects us to do the problem without using the counting formula (which is not introduced until section 9), though.

Exercise 6.7.11 According to Exercise 2.8.10, any subgroup $$G$$ of order 12 of $$S_4$$ must be normal, therefore it must be the kernel of some homomorphism $$\varphi : S_4 \to H$$ where $$H$$ is a group of order 2 and therefore has the form $$\{+1, -1\}$$. Let $$\sigma$$ be a 3-cycle in $$S_4$$. Then $$+1 = \varphi(1) = \varphi(\sigma^3) = \varphi(\sigma)^3$$, so $$\varphi(\sigma) = +1$$. But according to Exercise 2.4.11(b), the 3-cycles generate $$A_4$$. Therefore every even permutation in $$S_4$$ is mapped to $$+1$$; that is, $$A_4 \subseteq G$$. But $$A_4$$ has order 12, so $$G = A_4$$.