Brian Bi
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Miscellaneous Problems for Chapter 6

Problem 6.M.2

1. The identity function is obviously an automorphism of $$G$$. If $$\varphi \in \Aut G$$, then $$\varphi^{-1}$$ is also a bijection. Suppose $$a, b \in G$$, and let $$c = \varphi^{-1}(a), d = \varphi^{-1}(b)$$. Then $$a = \varphi(c), b = \varphi(d)$$, so $$ab = \varphi(cd)$$, so $$\varphi^{-1}(ab) = cd = \varphi^{-1}(a) \varphi^{-1}(b)$$, so $$\varphi^{-1}$$ preserves the operation of $$G$$. Thus, the inverse of an automorphism on $$G$$ is always an automorphism on $$G$$. Now consider two automorphisms $$\varphi_1, \varphi_2$$ of $$G$$. Then for every $$a, b \in G$$, $$\varphi_1 \circ \varphi_2(ab) = \varphi_1(\varphi_2(a)\varphi_2(b)) = \varphi_1(\varphi_2(a))\varphi_1(\varphi_2(b)) = (\varphi_1 \circ \varphi_2(a))(\varphi_1 \circ \varphi_2(b))$$, so $$\varphi_1 \circ \varphi_2$$ preserves the operation of $$G$$. And since $$\varphi_1, \varphi_2$$ are both bijections on the set $$G$$, so is their composition. Therefore $$\varphi_1 \circ \varphi_2$$ is an automorphism, and $$\Aut G$$ is closed under composition. These results together establish that $$\Aut G$$ is a group under the operation of composition.
2. Denote the operation of conjugation by $$g$$ by $$C_g$$. First, fix $$g$$ and consider the function $$C_g$$. For every $$h_1, h_2 \in H$$, we have that $$C_g(h_1 h_2) = gh_1 h_2 g^{-1} = (gh_1 g^{-1})(gh_2 g^{-1}) = C_g(h_1) C_g(h_2)$$. Therefore $$C_g$$ is a homomorphism for each $$g$$. It is also injective, because if $$C_g(h) = 1$$ then $$ghg^{-1} = 1$$, which we can rearrange to yield $$h = 1$$. Therefore $$C_g$$ is an automorphism for each $$g$$. So far we know that $$\varphi$$ which sends $$g$$ to $$C_g$$ is a function from $$G$$ to $$\Aut G$$. We now need to show that it is a homomorphism. Let $$g, g' \in G$$. Then $$C_g \circ C_{g'}$$ is the operation $$h \mapsto gg'hg'^{-1}g^{-1}$$, which is $$C_{gg'}(h)$$, so $$\varphi(gg') = \varphi(g)\varphi(g')$$ for all $$g, g' \in G$$. So $$\varphi$$ is indeed a homomorphism. The kernel consists of those elements of $$G$$ by which conjugation is the identity operation, that is, $$g$$ such that $$ghg^{-1} = h$$ for all $$h \in G$$. Such $$g$$ are the elements that commute with all elements of $$G$$, that is, $$\ker \varphi$$ is the centre of $$G$$.
3. It is clear enough that the inner automorphisms form a group, since they are the image of the homomorphism $$\varphi$$. Suppose $$i$$ is an inner automorphism of $$G$$, which conjugates by $$g \in G$$. Suppose $$a$$ is some arbitrary automorphism of $$G$$. Let $$h \in G$$. Then $$a \circ i \circ a^{-1}(h) = a(ga^{-1}(h)g^{-1}) = a(g)a(a^{-1}(h))a(g^{-1}) = a(g)h(a(g))^{-1}$$. That is, the automorphism $$aia^{-1}$$ is conjugation by $$a(g)$$, which is another inner automorphism. Since this holds for all $$i, a$$, the inner automorphism group is normal.

Problem 6.M.3

1. We previously argued (see Exercise 2.6.10(a)) that the automorphisms of $$C_n$$ are precisely the maps $$\varphi_k$$ defined by $$\varphi_k(x) = kx$$ for each $$k$$ relatively prime to $$n$$, where we restrict $$k$$ to be between 1 and $$n-1$$ to avoid duplicates. It's easy to see that the composition of such automorphisms is given by $$\varphi_k \circ \varphi_{k'} = \varphi_{kk'}$$. In the case $$n = 4$$, there are two automorphisms, $$\varphi_1$$ and $$\varphi_3$$, so the group of these two automorphisms is isomorphic to $$C_2$$.
2. Here again there are only two automorphisms since only 1 and 5 are relatively prime to 6. Again, the automorphism group is simply $$C_2$$.
3. The group $$C_2 \times C_2$$ consists of the identity and the elements $$x = (1, 0), y = (0, 1), xy = (1, 1)$$, with the operation being elementwise addition modulo 2. The three elements $$x, y, xy$$ all have order 2 and the product of any pair of them in either order equals the third. Any automorphism of $$C_2 \times C_2$$ must fix 1 so there are 6 candidate automorphisms, one for each permutation of $$\{x, y, xy\}$$. Each of these 6 permutations is a valid automorphism since the property of $$x, y, xy$$ having order 2 and having the product of each two be the third is preserved by any permutation of these three elements. Therefore, $$\Aut(C_2 \times C_2) \cong S_3$$.
4. The group $$D_4$$ is generated by $$\rho$$ and $$r$$ with the relations $$\rho^4 = 1, r^2 = 1, r\rho = \rho^{-1}r$$. The elements $$\rho$$ and $$\rho^3$$ are the only ones with order 4. Therefore if $$\varphi$$ is an automorphism of $$D_4$$, it must either map $$\rho$$ to $$\rho$$ or $$\rho$$ to $$\rho^{-1}$$. Since $$\rho, r$$ generate $$D_4$$, the automorphism is completely determined by $$\varphi(\rho)$$ and $$\varphi(r)$$. Since $$\varphi$$ maps rotations to rotations, there are 4 possible choices for $$\varphi(r)$$, so there are at most 8 possible automorphisms of $$D_4$$. Denote the automorphism that maps $$\rho$$ to $$\rho^i$$ with $$i \in \{1, -1\}$$ and $$r$$ to $$\rho^j r$$ with $$j \in \{0, 1, 2, 3\}$$ by $$\varphi_{ij}$$. Then $$\varphi_{ij}$$ maps $$r$$ to an element of order 2 and maps $$\rho$$ to an element of order 4, and $$\varphi_{ij}(r) \varphi_{ij}(\rho) = \rho^j r \rho^i = \rho^{j-i} r = (\rho^i)^{-1}(\rho^j r) = \varphi_{ij}(\rho)^{-1} \varphi_{ij}(r)$$, so $$\varphi_{ij}$$ preserves the defining relations of $$D_4$$. This establishes that each possible $$\varphi_{ij}$$ is indeed a valid automorphism of $$D_4$$ and there are no others.

Now $$P = \varphi_{1,1}$$ acts as a 4-cycle on the elements $$S = \{r, \rho r, \rho^2 r, \rho^3 r\}$$ while $$R = \varphi_{-1,0}$$ maps $$r \mapsto r, \rho r \mapsto \rho^3 r, \rho^2 r \mapsto \rho^2 r, \rho^3 r \mapsto \rho r$$. That is, $$P$$ and $$R$$ permute $$S$$ in the same way that $$\rho$$ and $$r$$ permute the four corners of a square. The action of $$D_4$$ on the vertices of the square is faithful, so there are 8 distinct permutations of the vertices of the square generated by the action of $$\rho$$ and $$r$$, which implies that all 8 permutations of $$S$$ are represented among the action of the subgroup of $$\Aut D_4$$ generated by $$P$$ and $$R$$. Since $$|\Aut D_4| = 8$$, this is the entire group $$\Aut D_4$$, and its action on $$S$$ is faithful. This implies that the law of composition of elements of $$\Aut D_4$$ is given by the composition of permutations by which elements of $$\Aut D_4$$ act. The same is true of $$D_4$$ itself, when regarded as acting on the 4 elements of $$S$$. Therefore $$\Aut D_4 \cong D_4$$.

5. Every automorphism of $$H$$ fixes $$-1$$, which is the only element of order 2. An automorphism therefore permutes the other 6 elements $$\pm i, \pm j, \pm k$$. We argue that $$\Aut H$$ is isomorphic to the group of rotational symmetries of a cube. To see this, label the faces of a cube with the symbols $$\pm i, \pm j, \pm k$$ so that $$\pm i$$ label opposite faces and similarly with $$\pm j$$ and $$\pm k$$, and the arrangement of the faces $$i, j, k$$ around their common vertex is counterclockwise. By inspection, we see that the product of any two faces is either $$-1$$ if they are opposite faces, or else is given by the unique third face such that the first factor, the second factor, and the product occur in counterclockwise order around their common vertex. Let $$\varphi$$ be an automorphism of $$H$$. Then $$\varphi$$ maps opposite faces to opposite faces (as opposite faces of the cube are labelled with inverse elements of $$H$$), so it maps the elements $$i, j, k$$ to three faces $$\varphi(i), \varphi(j), \varphi(k)$$ that meet at a common vertex and occur in counterclockwise order around that vertex. Now, each permutation of the faces of the cube with this property is induced by a unique rotational symmetry of the cube, namely the one that maps the common vertex of $$i, j, k$$ to the common vertex of $$\varphi(i), \varphi(j), \varphi(k)$$ (which preserves the counterclockwise orientation) and then rotates those three faces to achieve the desired final configuration. That is, to each automorphism of $$H$$ there corresponds a unique rotational symmetry of the cube. Conversely, if $$g$$ is some rotational symmetry of the cube, then the permutation it induces on the symbols $$\pm i, \pm j, \pm k$$ is an automorphism of $$H$$, because $$g$$ maps opposite faces to opposite faces (thus preserving inverses) and because if $$a, b \in H \setminus \{-1, 1\}$$ then the faces labelled with $$a, b, ab$$ on the cube will be mapped to three faces that also meet at a common vertex and occur in counterclockwise order, which therefore correspond to three elements in $$H \setminus \{-1, 1\}$$ with the third being the product of the first and second. (Note that any permutation of $$H \setminus \{-1, 1\}$$ preserves the products of an element from this set and either $$+1$$ or $$-1$$.) This establishes that the automorphisms of $$H$$ are in bijection with the elements of $$O$$. Obviously, the composition of two elements in $$O$$ permutes the elements of $$H \setminus \{-1, 1\}$$ in the same way as the two elements acting one after another, so $$\Aut H \cong O$$. (Note: an argument similar to that used in Exercise 6.12.3 can be used to show that $$O$$ acts faithfully on the 4 body diagonals of the cube, so alternatively $$\Aut H \cong S_4$$.)

Problem 6.M.4

1. We claim that the order of $$G_n$$ is $$n! 2^n$$. We will prove this by induction. The cases $$n = 1, n = 2$$ are easy and $$n = 2$$ can serve as a base case. Assume the order of $$G_{n-1}$$ is $$(n-1)!2^{n-1}$$. The group $$G_n$$ acts transitively on the $$2n$$ facets of $$\mathcal{C}_n$$. Let $$F$$ be the facet consisting of the vertices $$(1, \pm 1, \pm 1, \ldots, \pm 1)$$. The centre of $$F$$ is therefore $$(1, 0, \ldots, 0)$$, and the axis joining $$F$$ to the centre of $$\mathcal{C}_n$$ is the $$e_1$$ axis. Suppose $$g$$ is an isometry that fixes $$F$$ and therefore the centre of $$F$$; this automatically fixes the centre of the opposite facet $$(-1, 0, \ldots, 0)$$ as well. The matrix of $$g$$ is an orthogonal matrix whose first column is $$(1, 0, \ldots, 0)^t$$. Since the columns of an orthogonal matrix are orthonormal, this implies that the first row of $$g$$ is also $$(1, 0, \ldots, 0)$$, and the $$(n - 1) \times (n - 1)$$ matrix formed by deleting the first row and column can be any orthogonal matrix. That is, $$g$$ acts as an isometry $$g'$$ on the $$(n-1)$$-dimensional hyperplane perpendicular to $$e_1$$ and every such isometry can be extended to a unique isometry of $$\mathbb{R}^n$$ that fixes the centre of $$F$$. In order for $$g$$ to be a symmetry of $$\mathcal{C}_n$$, it is sufficient and necessary for $$g'$$ to be a symmetry of the intersection of $$\mathcal{C}_n$$ with the $$(n-1)$$-dimensional hyperplane perpendicular to $$e_1$$, which is the set $$\{(0, x_2, x_3, \ldots, x_n) \mid -1 \leq x_i \leq +1\}$$, and is therefore an $$(n-1)$$-dimensional hypercube. (It is sufficient because $$g$$ will then fix the set of all $$n$$ coordinate axes consisting of $$e_1$$ and the $$n-1$$ axes $$e_2, \ldots, e_n$$ which are obtained as the joining opposite facets of the $$(n-1)$$-dimensional hypercube, and these $$n$$ coordinate axes then determine the position of the $$n$$-dimensional hypercube.) By the induction hypothesis, there are $$(n-1)!2^{n-1}$$ choices for $$g'$$, thus the stabilizer of the centre of $$F$$ (and thus $$F$$ itself) has order $$(n-1)!2^{n-1}$$. The order of the entire $$\mathcal{C}_n$$ is therefore $$2n(n-1)!2^{n-1} = n!2^n$$.
2. The $$n!2^n$$ matrices that are obtained by taking some $$n \times n$$ permutation matrix and then changing the sign of some subset of the entries are all orthogonal, form a group, and all map vertices of $$\mathcal{C}_n$$ to vertices, so this group of matrices is precisely the group $$G_n$$. Such a matrix maps $$(1, 1, \ldots, 1)^t$$ to the element-wise sum of its $$n$$ column vectors, so an element of $$G_n$$ stabilizes this vertex iff all its entries are positive. That is, the stabilizer of this vertex is the group of $$n \times n$$ permutation matrices. Now the proper rotations in $$G_2$$ are generated by the matrix $P = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ which permutes the four column vectors $$(1, 1), (-1, 1), (-1, -1), (1, -1)$$ in a 4-cycle, and $$P$$ together with the orientation-reversing matrix $R = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ which permutes the same four vectors by transposing $$(1, -1)$$ and $$(-1, 1)$$. That is, $$P$$ and $$R$$ act on these four vectors in the same way as $$\rho, r \in D_4$$, respectively. By an argument analogous to that used in Problem 6.M.3(d), we conclude that $$G_2 \cong D_4$$, as we expect.

Problem 6.M.7

1. Denoting the generators of $$D_3$$ by $$\rho$$ and $$r$$ where $$\rho$$ is a 120 degree counterclockwise rotation, $$r$$ is a reflection that fixes vertex 1, and the vertices are labelled 1, 2, 3 in counterclockwise order:
 1 2 3 1 T T T $$\rho$$ F F F $$\rho^2$$ F F F $$r$$ T F F $$\rho r$$ F F T $$\rho^2 r$$ F T F
2. The number $$|G_s|$$ is just the number of true entries in the column of the table, so $$\sum_{s\in S} |G_s|$$ is the total number of true entries in the table. Similarly $$|S^g|$$ is the number of true entries in the row of the table, so $$\sum_{g \in G} |S^g|$$ is also the total number of true entries in the table.
3. Let $$\mathcal{O}$$ denote the set of orbits of $$S$$. Then $$\sum_{g \in G} |S^g| = \sum_{s \in S} |G_s| = \sum_{s\in S} \frac{|G|}{|O_s|} = |G| \sum_{s \in S} \frac{1}{|O_s|}$$, where we write $$O_s$$ for the orbit containing $$s$$. Since each element of $$S$$ is contained in exactly one orbit, we can rewrite this sum as $$|G| \sum_{O \in \mathcal{O}} \sum_{s \in O} \frac{1}{|O_s|} = |G| \sum_{O \in \mathcal{O}} \frac{|O|}{|O|} = |G||\mathcal{O}|$$.

Problem 6.M.8 Label the 8 edges of the octagon with the numbers 0, 1, 2, 3, 4, 5, 6, 7 in order. Let $$\rho$$ denote a rotation by angle $$\pi/4$$ and $$r$$ a reflection that fixes two vertices of the octagon. The identity element fixes all 70 colorings. The element $$\rho$$ doesn't fix any of the colorings (in order for it to do so, all 8 edges would have to have the same color). Likewise $$\rho^3, \rho^5, \rho^7$$, which also cycle each edge through the set of all 8 edges, don't fix any of the colorings. The colorings fixed by $$\rho^2$$ and $$\rho^6$$ are those such that alternating edges have the same color, of which there are two. The colorings fixed by $$\rho^4$$ are those where each edge has the same color as its opposite edge. Two pairs of opposite edges must therefore be white, so there are $$\binom{4}{2} = 6$$ elements fixed by $$\rho^4$$. The reflection $$r$$ exchanges four pairs of edges so it fixes those colorings where the two edges in each pair have the same color; again, two of these pairs must be white, so there are 6 elements fixed by $$r$$. The reflections $$\rho^2 r, \rho^4 r, \rho^6 r$$ perform the same kind of reflection, so each of these also fixes 6 colorings. Finally, the other four reflections $$\rho^k r$$ with $$k$$ odd each fix two sides and exchange the sides in three other pairs. A coloring fixed by such a reflection therefore has an even number of black sides among the 6 non-fixed sides, implying that the two fixed sides must have the same color. If the two fixed sides are black, we can choose one of the other three pairs to be black, and likewise if the two fixed sides are white then we can choose one of the other three pairs to be white. This implies that each such reflection fixes 6 colorings. In all, \begin{align*} \sum_{g \in G} |S^g| &= S^1 + 4S^\rho + 2S^{\rho^2} + S^{\rho^4} + 4S^r + 4S^{\rho r} \\ &= 70 + 4\times 0 + 2 \times 2 + 6 + 4\times 6 + 4\times 6 \\ &= 128 \end{align*} so the number of orbits is $$128 \divides |D_8| = 8$$. Representatives of these orbits are 0123, 0124, 0125, 0134, 0135, 0136, 0145, 0246, where each set of 4 digits represents a set of edges to be colored black.