Homework 2: Group Theory

1. Since \(ea = ae = a\), we have \(e \in C_{G}(a)\), so \(C_{G}(a)\) is nonempty. Let \(x, y \in C_{G}(a)\). Then \(xa = ax\) and \(ya = ay\). From \(ya = ay\) we obtain

   \[ y^{-1}a = ay^{-1}. \]

   Indeed, multiplying \(ya = ay\) on the left and right by \(y^{-1}\) gives this identity. Therefore

   \[ (xy^{-1})a = x(y^{-1}a) = x(ay^{-1}) = (xa)y^{-1} = (ax)y^{-1} = a(xy^{-1}). \]

   Thus \(xy^{-1} \in C_{G}(a)\). By the subgroup test, \(C_{G}(a) \le G\).

2. The six elements of \(S_{3}\) are

   \[ e, \ (1\,2), \ (1\,3), \ (2\,3), \ (1\,2\,3), \ (1\,3\,2). \]

   Certainly \(e\) and \((1\,2)\) commute with \((1\,2)\). We check that no other element does. For example,

   \[ (1\,2)(1\,3) = (1\,3\,2), \quad (1\,3)(1\,2) = (1\,2\,3), \]

   so \((1\,3)\) does not commute with \((1\,2)\). Next,

   \[ (1\,2)(2\,3) = (1\,2\,3), \quad (2\,3)(1\,2) = (1\,3\,2), \]

   so \((2\,3)\) does not commute with \((1\,2)\). For the two 3-cycles,

   \[ (1\,2)(1\,2\,3) = (2\,3), \quad (1\,2\,3)(1\,2) = (1\,3), \]

   and

   \[ (1\,2)(1\,3\,2) = (1\,3), \quad (1\,3\,2)(1\,2) = (2\,3). \]

   Thus neither 3-cycle commutes with \((1\,2)\). Hence

   \[ C_{S_{3}}((1\,2)) = \{e, (1\,2)\}. \]

Write \(x = a^{k}\). Since \(a\) has order 18, two powers \(a^{u}\) and \(a^{v}\) are equal exactly when 18 divides \(u - v\). Indeed, \(a^{u} = a^{v}\) precisely when \(a^{u-v} = e\), and the powers of \(a\) equal to \(e\) are exactly those whose exponents are multiples of 18.

1. The equation \(x^{3} = e\) becomes \(a^{3k} = e\). Therefore 18 divides \(3k\). This means that \(3k = 18q\) for some \(q \in \mathbb{Z}\), and hence \(k = 6q\). Thus 6 divides \(k\). Modulo 18, the possible values of \(k\) are \(0, 6\), and \(12\). Hence

   \[ x \in \{e, a^{6}, a^{12}\}. \]

2. The equation \(x^{3} = a^{6}\) is equivalent to

   \[ 3k \equiv 6 \pmod{18}. \]

   Dividing by 3 gives

   \[ k \equiv 2 \pmod{6}. \]

   Modulo 18, the solutions are \(k \equiv 2, 8, 14\). Thus

   \[ x \in \{a^{2}, a^{8}, a^{14}\}. \]

3. The equation \(x^{4} = a^{6}\) is equivalent to

   \[ 4k \equiv 6 \pmod{18}. \]

   Dividing by 2 gives \(2k \equiv 3 \pmod{9}\). Since 5 is an inverse of 2 modulo 9, this is equivalent to

   \[ k \equiv 15 \equiv 6 \pmod{9}. \]

   Modulo 18, the two solutions are \(k \equiv 6, 15\). Hence

   \[ x \in \{a^{6}, a^{15}\}. \]

We have

1. Hence

   \[ HK = \{e, (1\,2), (2\,3), (1\,2)(2\,3)\} = \{e, (1\,2), (2\,3), (1\,2\,3)\}. \]

   For the reverse product,

   \[ KH = \{e, (1\,2), (2\,3), (2\,3)(1\,2)\} = \{e, (1\,2), (2\,3), (1\,3\,2)\}. \]

2. Since \((1\,2\,3) \in HK\) but \((1\,2\,3) \notin KH\), the two sets are not equal.

3. Both \((1\,2)\) and \((2\,3)\) belong to \(HK\). If \(HK\) were a subgroup, it would contain their product in the opposite order,

   \[ (2\,3)(1\,2) = (1\,3\,2). \]

   But \((1\,3\,2) \notin HK\) by part (a). Thus \(HK\) is not closed under multiplication and is therefore not a subgroup of \(S_{3}\).

From \(srs = r^{-1}\) one obtains \(sr^{i}s = r^{-i}\) for every integer \(i\).

1. Therefore

   \[ (sr^{i})(sr^{j}) = sr^{i}s r^{j} = r^{-i}r^{j} = r^{j-i}. \]

2. Taking \(j = i\) in part (a),

   \[ (sr^{i})^{2} = r^{i-i} = e. \]

   Since \(sr^{i}\) is a reflection and hence is not the identity, its order is exactly 2.

3. By part (a),

   \[ (sr^{i})(sr^{j}) = r^{j-i}, \quad (sr^{j})(sr^{i}) = r^{i-j}. \]

   Thus the two reflections commute exactly when

   \[ r^{j-i} = r^{i-j}, \]

   or equivalently when

   \[ r^{2(i-j)} = e. \]

   Since \(r\) has order \(n\), this is equivalent to

   \[ n \mid 2(i-j). \]

   If \(n\) is odd, this forces \(n \mid (i-j)\), so the two reflections are the same. If \(n\) is even, write \(n = 2m\). Then \(n \mid 2(i-j)\) is equivalent to \(m \mid (i-j)\). Modulo \(n\), this means either

   \[ i \equiv j \pmod{n} \quad \text{or} \quad i \equiv j + \frac{n}{2} \pmod{n}. \]

   Thus for even \(n\), two distinct reflections commute precisely when their indices differ by \(n/2\) modulo \(n\).

1. Suppose \([m]_{12} = [m']_{12}\). Then \(12 \mid (m - m')\), so \(m - m' = 12q\) for some \(q \in \mathbb{Z}\). Hence

   \[ 3m - 3m' = 36q, \]

   which is divisible by 18. Therefore

   \[ [3m]_{18} = [3m']_{18}. \]

   Thus \(\varphi\) is well defined.

2. For \([m]_{12}, [n]_{12} \in \mathbb{Z}/12\mathbb{Z}\),

   \[ \varphi([m]_{12} + [n]_{12}) = \varphi([m+n]_{12}) = [3(m+n)]_{18} = [3m]_{18} + [3n]_{18}. \]

   Hence \(\varphi\) is a homomorphism.

3. An element \([m]_{12}\) lies in \(\operatorname{Ker} \varphi\) precisely when \(\varphi([m]_{12}) = [0]_{18}\). By the definition of \(\varphi\), this means that \([3m]_{18} = [0]_{18}\), so \(18\) divides \(3m\). Equivalently, \(6\) divides \(m\). Among the residue classes modulo \(12\), this occurs only for \([0]_{12}\) and \([6]_{12}\). Thus

   \[ \operatorname{Ker} \varphi = \{[0]_{12}, [6]_{12}\}. \]

   The possible values of \([3m]_{18}\) are precisely

   \[ [0]_{18}, [3]_{18}, [6]_{18}, [9]_{18}, [12]_{18}, [15]_{18}. \]

   Hence

   \[ \operatorname{im} \varphi = \{[0]_{18}, [3]_{18}, [6]_{18}, [9]_{18}, [12]_{18}, [15]_{18}\}. \]

4. The map is not injective because

   \[ [0]_{12} \ne [6]_{12} \quad \text{but} \quad \varphi([0]_{12}) = \varphi([6]_{12}) = [0]_{18}. \]

   It is not surjective because, for example, \([1]_{18}\) is not in the image.

1. The set \(H\) is nonempty because \((0,0) \in H\). Let \((m, n), (r, s) \in H\). Then \(3 \mid (m - n)\) and \(3 \mid (r - s)\). Hence

   \[ (m - r) - (n - s) = (m - n) - (r - s) \]

   is divisible by 3. Thus

   \[ (m, n) - (r, s) = (m - r, n - s) \in H. \]

   By the subgroup test for the additive group \(\mathbb{Z}^{2}\), \(H \le \mathbb{Z}^{2}\).

2. Let \(K = \langle (1, 1), (3, 0) \rangle\). Every element of \(K\) has the form

   \[ a(1, 1) + b(3, 0) = (a + 3b, a) \]

   for some \(a, b \in \mathbb{Z}\). Its two coordinates differ by \(3b\), so \(K \subseteq H\). Conversely, let \((m, n) \in H\). Since \(3 \mid (m - n)\), there is \(b \in \mathbb{Z}\) such that \(m - n = 3b\). Hence

   \[ (m, n) = n(1, 1) + b(3, 0) \in K. \]

   Therefore \(H = K\).

3. We compute the differences of the coordinates:

   \[ 7 - 1 = 6, \quad 5 - (-1) = 6, \quad 4 - 2 = 2. \]

   The first two are divisible by \(3\), while the third is not. Thus \((7, 1), (5, -1) \in H\) and \((4, 2) \notin H\).

1. The identity preserves \(\{1, 2\}\), so \(H\) is nonempty. If \(\sigma, \tau \in H\), then \(\tau^{-1}\) also preserves \(\{1, 2\}\) setwise, because \(\tau\) restricts to a bijection of \(\{1, 2\}\) onto itself. Therefore

   \[ (\sigma\tau^{-1})(\{1, 2\}) = \sigma(\tau^{-1}(\{1, 2\})) = \sigma(\{1, 2\}) = \{1, 2\}. \]

   Thus \(\sigma\tau^{-1} \in H\), and the subgroup test gives \(H \le S_{4}\).

2. A permutation in \(H\) may independently interchange \(1, 2\) and interchange \(3, 4\). Thus

   \[ H = \{e, (1\,2), (3\,4), (1\,2)(3\,4)\}. \]

3. Write \(C_{2} = \{0, 1\}\) additively. Define

   \[ \Phi: C_{2} \times C_{2} \longrightarrow H, \quad \Phi(a, b) = (1\,2)^{a}(3\,4)^{b}. \]

   The two transpositions are disjoint and hence commute, so

   \[ \Phi(a, b)\Phi(c, d) = (1\,2)^{a+c}(3\,4)^{b+d} = \Phi(a+c, b+d), \]

   where exponents are read modulo 2. Thus \(\Phi\) is a homomorphism. The four pairs

   \[ (0, 0), \ (1, 0), \ (0, 1), \ (1, 1) \]

   map to the four distinct elements listed in part (b), so \(\Phi\) is bijective. Hence it is an isomorphism.

1. Since \(e^{m} = e\), the set \(G[m]\) is nonempty. Let \(x, y \in G[m]\). Because \(G\) is abelian,

   \[ (xy^{-1})^{m} = x^{m}(y^{-1})^{m} = x^{m}(y^{m})^{-1} = e e^{-1} = e. \]

   Hence \(xy^{-1} \in G[m]\). By the subgroup test, \(G[m] \le G\).

2. In \(S_{3}\), both \((1\,2)\) and \((2\,3)\) satisfy \(\sigma^{2} = e\). However, using right-to-left composition,

   \[ (1\,2)(2\,3) = (1\,2\,3), \]

   and

   \[ (1\,2\,3)^{2} = (1\,3\,2) \ne e. \]

   Thus the set \(\{\sigma \in S_{3} : \sigma^{2} = e\}\) is not closed under multiplication and therefore is not a subgroup.

1. The identity matrix lies in \(SL_{2}(\mathbb{Z})\). Let \(A, B \in SL_{2}(\mathbb{Z})\). Then

   \[ \det(AB^{-1}) = (\det A)(\det B)^{-1} = 1 \cdot 1^{-1} = 1. \]

   Thus \(AB^{-1} \in SL_{2}(\mathbb{Z})\). By the subgroup test, \(SL_{2}(\mathbb{Z}) \le GL_{2}(\mathbb{Z})\).

2. A subgroup must contain the identity, but \(\det I = 1\), so \(I \notin T\). Thus \(T\) cannot be a subgroup. It also fails closure under multiplication: if \(A, B \in T\), then \(\det(AB) = (-1)(-1) = 1\), so \(AB \notin T\).

3. For example,

   \[ A = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \]

   both have determinant \(-1\), while

   \[ AB = -I \]

   has determinant \(1\).

Suppose first that \(f\) is a homomorphism. The graph is nonempty because \((e_{G}, f(e_{G})) = (e_{G}, e_{K}) \in \Gamma_{f}\). If \((g, f(g)), (h, f(h)) \in \Gamma_{f}\), then

where the last equality uses the homomorphism property. Hence this element lies in \(\Gamma_{f}\), so \(\Gamma_{f} \le G \times K\).

Conversely, suppose \(\Gamma_{f}\) is a subgroup. Since the identity of \(G \times K\) is \((e_{G}, e_{K})\), membership of the identity in the graph gives \(f(e_{G}) = e_{K}\). For \(g, h \in G\), both \((g, f(g))\) and \((h, f(h))\) lie in the graph, so their product lies in the graph:

By definition of the graph, the unique element of \(\Gamma_{f}\) whose first coordinate is \(gh\) is \((gh, f(gh))\). Therefore

Thus \(f\) is a homomorphism.

1. An integer \(x\) belongs to \(m\mathbb{Z} \cap n\mathbb{Z}\) exactly when \(m \mid x\) and \(n \mid x\). This is equivalent to saying that \(x\) is a multiple of \(\operatorname{lcm}(m, n)\). Hence

   \[ m\mathbb{Z} \cap n\mathbb{Z} = \operatorname{lcm}(m, n)\mathbb{Z}. \]

2. Put \(d = \gcd(m, n)\). Since \(d \mid m\) and \(d \mid n\), both \(m\mathbb{Z}\) and \(n\mathbb{Z}\) are contained in \(d\mathbb{Z}\). Therefore

   \[ \langle m\mathbb{Z} \cup n\mathbb{Z} \rangle \subseteq d\mathbb{Z}. \]

   By Bézout's identity, there exist \(u, v \in \mathbb{Z}\) such that

   \[ d = um + vn. \]

   Since the subgroup generated by \(m\mathbb{Z} \cup n\mathbb{Z}\) contains both \(m\) and \(n\), it contains \(um + vn = d\), and hence it contains every multiple of \(d\). Thus

   \[ d\mathbb{Z} \subseteq \langle m\mathbb{Z} \cup n\mathbb{Z} \rangle. \]

   The two inclusions give equality.

1. Reducing exponents modulo 36,

   \[ H = \{a^{4j} : 0 \le j \le 8\} = \{e, a^{4}, a^{8}, a^{12}, a^{16}, a^{20}, a^{24}, a^{28}, a^{32}\}, \]

   while

   \[ K = \{a^{6j} : 0 \le j \le 5\} = \{e, a^{6}, a^{12}, a^{18}, a^{24}, a^{30}\}. \]

2. The common exponents modulo 36 are 0, 12, 24. Hence

   \[ H \cap K = \{e, a^{12}, a^{24}\} = \langle a^{12} \rangle. \]

3. Since \(a^{4}, a^{6} \in \langle H \cup K \rangle\), the subgroup also contains

   \[ a^{6}(a^{4})^{-1} = a^{2}. \]

   Thus \(\langle a^{2} \rangle \subseteq \langle H \cup K \rangle\). Conversely, every power of \(a^{4}\) and every power of \(a^{6}\) is an even power of \(a\), so \(H \cup K \subseteq \langle a^{2} \rangle\). Therefore

   \[ \langle H \cup K \rangle = \langle a^{2} \rangle. \]

1. Since every element of \(G\) is a power of \(a\), the value of \(\varphi\) is determined by \(\varphi(a)\). There is a unique \(k \in \mathbb{Z}\) such that \(\varphi(a) = a^{k}\), because the powers of \(a\) are distinct. Then

   \[ \varphi(a^{m}) = \varphi(a)^{m} = (a^{k})^{m} = a^{km} \]

   for every \(m \in \mathbb{Z}\).

2. If \(k = 0\), then \(\varphi\) is the trivial homomorphism and is not injective unless \(G\) is trivial, which it is not. If \(k \ne 0\) and \(\varphi(a^{m}) = \varphi(a^{n})\), then \(a^{km} = a^{kn}\). Since \(a\) has infinite order, \(km = kn\), and because \(k \ne 0\), \(m = n\). Hence \(\varphi\) is injective exactly when \(k \ne 0\).

3. The image is

   \[ \{a^{km} : m \in \mathbb{Z}\} = \langle a^{k} \rangle. \]

   This equals all of \(G\) precisely when \(a\) itself is a power of \(a^{k}\), i.e. when there is \(m \in \mathbb{Z}\) with \(km = 1\). This occurs exactly for \(k = 1\) or \(k = -1\).

4. An automorphism is both injective and surjective, so the only possibilities are \(a^{m} \mapsto a^{m}\) and \(a^{m} \mapsto a^{-m}\).

Set \(d_{r} = \gcd(n, r)\). By Bézout's identity there exist \(u, v \in \mathbb{Z}\) such that

Therefore

so \(a^{d_{r}} \in \langle a^{r} \rangle\). Since \(d_{r} \mid r\), write \(r = d_{r}t\); then \(a^{r} = (a^{d_{r}})^{t}\), so \(a^{r} \in \langle a^{d_{r}} \rangle\). Hence

Similarly,

Thus if \(d_{r} = d_{s}\), the two subgroups are equal.

Conversely, suppose \(\langle a^{r} \rangle = \langle a^{s} \rangle\). From the first part of the argument,

so \(a^{d_{r}} = a^{st}\) for some \(t \in \mathbb{Z}\). Hence

Because \(d_{s}\) divides both \(n\) and \(s\), it follows that \(d_{s} \mid d_{r}\). Interchanging \(r\) and \(s\) gives \(d_{r} \mid d_{s}\). Since both are positive, \(d_{r} = d_{s}\).

1. The subgroup is

   \[ \langle a^{6} \rangle = \{e, a^{6}, a^{12}, a^{18}\}. \]

   The elements \(a^{6}\) and \(a^{18} = (a^{6})^{3}\) generate all four elements. The identity generates only \(\{e\}\), while \(a^{12}\) has square \(e\) and generates only \(\{e, a^{12}\}\). Hence the required elements are

   \[ a^{6} \quad \text{and} \quad a^{18}. \]

2. Any generator of \(\langle a^{6} \rangle\) must itself belong to that subgroup, so the four elements listed above exhaust all possibilities. The direct computations in part (a) show that only \(a^{6}\) and \(a^{18}\) generate it.

3. We have

   \[ \langle a^{4} \rangle = \{e, a^{4}, a^{8}, a^{12}, a^{16}, a^{20}\}. \]

   Writing \(b = a^{4}\), this is the cyclic group

   \[ \{e, b, b^{2}, b^{3}, b^{4}, b^{5}\}. \]

   The powers \(b^{j}\) generate the whole set exactly for \(j=1\) and \(j=5\): indeed, \(b, b^{5}\) run through all six powers, while \(b^{2}, b^{4}\) repeat after three powers, \(b^{3}\) repeats after two, and \(e\) after one. Thus

   \[ y = a^{4} \quad \text{or} \quad y = a^{20}. \]

1. Conjugation relabels the entries of a cycle. Since \(c^{i-1}(1) = i\) and \(c^{i-1}(2) = i+1\), we obtain

   \[ c^{i-1}(1\,2)c^{-(i-1)} = (i\,i+1). \]

2. Let \(i < j\). Consider the product

   \[ (i\,i+1)(i+1\,i+2)\dots(j-2\,j-1)(j-1\,j)(j-2\,j-1)\dots(i\,i+1). \]

   Composing from right to left, the point \(i\) moves successively through \(i+1, i+2, \dots, j\) so it is sent to \(j\). The point \(j\) is sent first to \(j-1\) and then successively back through \(j-2, \dots, i\) so it is sent to \(i\). Every point outside \(\{i, \dots, j\}\) is fixed, and each intermediate point \(i < k < j\) is moved once to a neighboring point and then moved back. Hence this product is exactly \((i\,j)\). Therefore every transposition is a product of adjacent transpositions.

3. Every permutation is a product of transpositions. By part (b), every transposition is a product of adjacent transpositions, and by part (a), every adjacent transposition belongs to \(\langle c, t \rangle\). Hence every element of \(S_{n}\) lies in \(\langle c, t \rangle\). Conversely, since \(c, t \in S_{n}\) and \(S_{n}\) is a subgroup containing both generators, every element of \(\langle c, t \rangle\) lies in \(S_{n}\). Therefore \(S_{n} = \langle c, t \rangle\).

1. Every \(\sigma \in W\) induces a permutation of the two blocks of \(\mathcal{P}\). The identity does so. If \(\sigma, \tau \in W\), then \(\tau^{-1}\) also permutes the two blocks, and therefore so does \(\sigma\tau^{-1}\). Hence \(W \le S_{4}\) by the subgroup test.

2. If the two blocks are preserved individually, one may independently swap the two points in each block, giving

   \[ e, \ (1\,2), \ (3\,4), \ (1\,2)(3\,4). \]

   If the blocks are interchanged, the restriction of \(\sigma\) to \(\{1, 2\}\) can be either of the two bijections from \(\{1, 2\}\) to \(\{3, 4\}\). Independently, the restriction of \(\sigma\) to \(\{3, 4\}\) can be either of the two bijections from \(\{3, 4\}\) to \(\{1, 2\}\). Thus there are four block-interchanging permutations:

   \[ (1\,3)(2\,4), \ (1\,4)(2\,3), \ (1\,3\,2\,4), \ (1\,4\,2\,3). \]

   These are the eight elements of \(W\).

3. The cycle \(r\) interchanges the two blocks and \(s\) preserves them, so both lie in \(W\). Since \(r\) is a 4-cycle, \(r^{4} = e\). Since \(s\) is a transposition, \(s^{2} = e\). Conjugating \(r\) by \(s\) relabels its entries:

   \[ srs = (s(1)\,s(3)\,s(2)\,s(4)) = (2\,3\,1\,4) = (1\,4\,2\,3) = r^{-1}. \]

4. We compute the powers of \(r\) and then compose each of them on the left with \(s\) (permutations are composed from right to left):

   \[\begin{aligned} r^{0} &= e, & r &= (1\,3\,2\,4), & r^{2} &= (1\,2)(3\,4), & r^{3} &= (1\,4\,2\,3), \\ s &= (1\,2), & sr &= (1\,3)(2\,4), & sr^{2} &= (3\,4), & sr^{3} &= (1\,4)(2\,3). \end{aligned}\]

   For example, \(sr\) sends 1 to 3, 3 to 1, 2 to 4, and 4 to 2, so \(sr = (1\,3)(2\,4)\). The remaining products are obtained in the same way. These eight permutations are precisely the elements listed in part (b). Hence

   \[ W = \{e, r, r^{2}, r^{3}, s, sr, sr^{2}, sr^{3}\}. \]

1. We have

   \[ sr^{j}(v_{k}) = s(v_{k+j}) = v_{-(k+j)}. \]

   Thus \(sr^{j}(v_{k}) = v_{k}\) exactly when \(-k-j \equiv k \pmod{n}\), i.e. \(2k \equiv -j \pmod{n}\).

2. If \(n\) is odd, \(2\) is invertible modulo \(n\). Hence for every \(j\) there is exactly one residue class \(k\) satisfying \(2k \equiv -j \pmod{n}\). Thus each reflection fixes exactly one vertex.

3. Write \(n = 2m\). The congruence \(2k \equiv -j \pmod{2m}\) has no solution if \(j\) is odd. If \(j = 2q\), it is equivalent to \(k \equiv -q \pmod{m}\), which gives exactly two residue classes modulo \(2m\), namely \(-q\) and \(-q+m\). Hence the reflection fixes two vertices for even \(j\) and none for odd \(j\).

4. When \(j\) is even, the reflection axis passes through two opposite vertices. When \(j\) is odd, it passes through the midpoints of a pair of opposite edges, so it fixes no vertex.

1. Repeated use of \(srs = r^{-1}\) gives

   \[ r^{k}sr^{-k} = sr^{-2k} \]

   for every integer \(k\). Therefore

   \[ r^{k}(sr^{j})r^{-k} = (r^{k}sr^{-k})r^{j} = sr^{-2k}r^{j} = sr^{j-2k}. \]

2. Since \(s^{-1} = s\), the defining relation gives

   \[ s(sr^{j})s^{-1} = sr^{-j}. \]

3. If \(n\) is odd, multiplication by \(2\) is a bijection on \(\mathbb{Z}/n\mathbb{Z}\). Given \(j, l\), choose \(k\) such that \(j - 2k \equiv l \pmod{n}\). Part (a) then shows that \(sr^{j}\) is conjugate to \(sr^{l}\).

4. Suppose \(n\) is even. Parts (a) and (b) show that conjugation changes the index \(j\) either to \(j - 2k\) or to \(-j\); in either case the parity of \(j\) is preserved. Hence an even-index reflection cannot be conjugate to an odd-index reflection. Conversely, if \(j\) and \(l\) have the same parity, then \(j - l\) is even, so the congruence \(2k \equiv j - l \pmod{n}\) has a solution. Part (a) then conjugates \(sr^{j}\) to \(sr^{l}\). Thus the even-index reflections form one conjugacy class, represented by \(s\), and the odd-index reflections form the other, represented by \(sr\).

1. We have \(r^{2} \ne e\) and \((r^{2})^{2} = e\), so \(r^{2}\) has order \(2\). Every reflection \(sr^{k}\) satisfies

   \[ (sr^{k})^{2} = sr^{k}sr^{k} = r^{-k}r^{k} = e. \]

   Thus the five elements

   \[ r^{2}, \ s, \ sr, \ sr^{2}, \ sr^{3} \]

   have order \(2\). The remaining nonidentity rotations \(r\) and \(r^{3}\) have order \(4\).

2. In \(Q_{8}\), \((-1)^{2} = 1\), so \(-1\) has order \(2\). On the other hand,

   \[ i^{2} = j^{2} = k^{2} = -1, \]

   so each of \(\pm i, \pm j, \pm k\) has order \(4\). Thus \(-1\) is the unique element of order \(2\) in \(Q_{8}\).

3. We first justify the fact about element orders that we shall use. Suppose \(\varphi: G \rightarrow H\) is an isomorphism and \(g \in G\). If \(g^{m} = e_{G}\), then

   \[ \varphi(g)^{m} = \varphi(g^{m}) = \varphi(e_{G}) = e_{H}. \]

   Conversely, if \(\varphi(g)^{m} = e_{H}\), then

   \[ \varphi(g^{m}) = \varphi(g)^{m} = e_{H} = \varphi(e_{G}). \]

   Since \(\varphi\) is injective, \(g^{m} = e_{G}\). Thus \(g\) and \(\varphi(g)\) have the same order. Because an isomorphism is bijective, it therefore gives a bijection between the elements of order \(2\) in the two groups.

   By parts (a) and (b), \(D_{4}\) has five elements of order \(2\), whereas \(Q_{8}\) has only one. Hence no isomorphism \(D_{4} \rightarrow Q_{8}\) can exist.

1. Reflexivity follows from \(f(x) = f(x)\). If \(f(x) = f(y)\), then \(f(y) = f(x)\), giving symmetry. If \(f(x) = f(y)\) and \(f(y) = f(z)\), then \(f(x) = f(z)\), giving transitivity. Hence \(\sim\) is an equivalence relation.

2. By definition,

   \[ [x] = \{y \in X : f(y) = f(x)\} = f^{-1}(\{f(x)\}). \]

3. For \(m \in \mathbb{Z}\), the equation \(n^{2} = m^{2}\) is equivalent to \(n = \pm m\). Thus

   \[ [0] = \{0\}, \quad [m] = \{m, -m\} \quad (m \ne 0). \]

   The partition is therefore

   \[ \{\{0\}, \{1, -1\}, \{2, -2\}, \{3, -3\}, \dots\}. \]

1. Equality of the integer \(a+b\) is reflexive, symmetric, and transitive, so \(\sim\) is an equivalence relation. Explicitly, \((a,b) \sim (a,b)\); if \(a+b = c+d\), then \(c+d = a+b\); and if \(a+b = c+d\) and \(c+d = p+q\), then \(a+b = p+q\).

2. We have

   \[ [(a, b)] = \{(x, y) \in \mathbb{Z}^{2} : x + y = a + b\}. \]

   Equivalently,

   \[ [(a, b)] = \{(t, a + b - t) : t \in \mathbb{Z}\}. \]

3. Each class consists of the lattice points on a line of slope \(-1\) having an integer intercept. Distinct values of \(a+b\) give disjoint lines, and every lattice point lies on exactly one such line. Hence these classes partition \(\mathbb{Z}^{2}\).

Definition (left and right cosets). Let \(G\) be a group, let \(H \le G\), and let \(g \in G\). The left coset of \(H\) determined by \(g\) is

and the right coset of \(H\) determined by \(g\) is

1. If \(\tau \in \sigma H\), then \(\tau = \sigma h\) for some \(h \in H\). Hence

   \[ \tau(1) = \sigma(h(1)) = \sigma(1). \]

   Conversely, suppose \(\tau(1) = \sigma(1)\). Then

   \[ (\sigma^{-1}\tau)(1) = \sigma^{-1}(\tau(1)) = \sigma^{-1}(\sigma(1)) = 1, \]

   so \(\sigma^{-1}\tau \in H\). Therefore \(\tau \in \sigma H\).

2. By part (a), a left coset is determined exactly by the value \(\sigma(1) \in \{1, 2, 3, 4\}\). Each value occurs, so there are exactly four cosets.

3. We may take

   \[ e, \ (1\,2), \ (1\,3), \ (1\,4) \]

   as representatives. Thus the partition is

   \[ S_{4} = \{\tau : \tau(1) = 1\} \sqcup \{\tau : \tau(1) = 2\} \sqcup \{\tau : \tau(1) = 3\} \sqcup \{\tau : \tau(1) = 4\}. \]

   Each set is the corresponding left coset of \(H\).

1. Since \(H = \{e, s\}\),

   \[\begin{aligned} H &= \{e, s\}, \\ rH &= \{r, sr^{3}\}, \\ r^{2}H &= \{r^{2}, sr^{2}\}, \\ r^{3}H &= \{r^{3}, sr\}. \end{aligned}\]

   These four sets are pairwise disjoint and together contain all eight displayed elements of \(D_{4}\), so they are all the left cosets.

2. Directly from the definition of a right coset,

   \[\begin{aligned} H &= \{e, s\}, \\ Hr &= \{r, sr\}, \\ Hr^{2} &= \{r^{2}, sr^{2}\}, \\ Hr^{3} &= \{r^{3}, sr^{3}\}. \end{aligned}\]

3. For example,

   \[ rH = \{r, sr^{3}\} \quad \text{whereas} \quad Hr = \{r, sr\}. \]

   Since \(sr^{3} \ne sr\), these cosets are different.

Let \(y \in aH \cap bK\). Since \(x, y \in aH\), there are \(h_{0}, h_{1} \in H\) such that \(x = ah_{0}\) and \(y = ah_{1}\). Hence

Similarly, since \(x, y \in bK\), we have \(x^{-1}y \in K\). Thus \(x^{-1}y \in H \cap K\), and therefore

This proves

Conversely, let \(z \in H \cap K\). Since \(x \in aH\), multiplication on the right by \(z \in H\) keeps us in the same left coset, so \(xz \in aH\). Likewise, \(x \in bK\) and \(z \in K\) imply \(xz \in bK\). Hence

Thus equality holds.

If \(x' \in aH \cap bK\) is another choice, the same argument gives

Hence both expressions are equal to the same intersection, so the formula is independent of the chosen point.

1. For \((m, n), (r, s) \in \mathbb{Z}^{2}\),

   \[ \varphi((m, n) + (r, s)) = 4(m + r) + 6(n + s) = (4m + 6n) + (4r + 6s), \]

   so \(\varphi\) is a homomorphism.

2. Every value \(4m + 6n\) is even, so \(\operatorname{im} \varphi \subseteq 2\mathbb{Z}\). Conversely,

   \[ \varphi(-1, 1) = 2, \]

   so every multiple of \(2\) lies in the image. Hence

   \[ \operatorname{im} \varphi = 2\mathbb{Z}. \]

3. Let \((m, n) \in \mathbb{Z}^{2}\). This element lies in \(\operatorname{Ker} \varphi\) precisely when

   \[ 4m + 6n = 0. \]

   Dividing this equation by \(2\) gives

   \[ 2m + 3n = 0. \]

   Thus \(2m = -3n\). Since the left-hand side is even, \(3n\) is even. Because \(3\) is odd, \(n\) must be even, so write \(n = 2t\) for some \(t \in \mathbb{Z}\). Substituting into \(2m + 3n = 0\) gives \(m = -3t\). Therefore

   \[ \operatorname{Ker} \varphi = \{t(-3, 2) : t \in \mathbb{Z}\} = \langle (-3, 2) \rangle. \]

1. For \(z, w \in \mathbb{C}^{\times}\),

   \[ \rho(zw) = |zw| = |z||w| = \rho(z)\rho(w), \]

   so \(\rho\) is a homomorphism.

2. The identity of \(\mathbb{R}_{>0}\) is \(1\), hence

   \[ \operatorname{Ker} \rho = \{z \in \mathbb{C}^{\times} : |z| = 1\}, \]

   the unit circle.

3. If \(r > 0\), then the positive real number \(z = r \in \mathbb{C}^{\times}\) satisfies \(\rho(z) = r\). Hence \(\rho\) is surjective.

4. It is not injective: for example,

   \[ \rho(1) = 1 = \rho(-1) \]

   although \(1 \ne -1\). Hence it is not an isomorphism.

1. Using the relation \(srs = r^{-1}\), the four possible products, written again in the normal forms \(r^{k}\) and \(sr^{k}\), are

   \[\begin{aligned} r^{i}r^{j} &= r^{i+j} \\ r^{i}(sr^{j}) &= sr^{j-i} \\ (sr^{i})r^{j} &= sr^{i+j} \\ (sr^{i})(sr^{j}) &= r^{j-i}. \end{aligned}\]

   Thus a product is a rotation exactly when the two factors are both rotations or both reflections, and it is a reflection exactly when precisely one factor is a reflection. This is exactly the multiplication rule for their signs in \(\{1, -1\}\). Hence

   \[ \epsilon(xy) = \epsilon(x)\epsilon(y) \]

   for all \(x, y \in D_{n}\).

2. The elements mapped to \(1\) are precisely the rotations, so

   \[ \operatorname{Ker} \epsilon = \langle r \rangle. \]

   Both \(1\) and \(-1\) occur as values, for example on \(e\) and \(s\), so

   \[ \operatorname{im} \epsilon = \{1, -1\}. \]

3. The map is surjective. It is not injective for \(n \ge 3\), since \(e \ne r\) but

   \[ \epsilon(e) = \epsilon(r) = 1. \]

1. For \(x, y \in G\),

   \[ c_{g}(xy) = gxyg^{-1} = (gxg^{-1})(gyg^{-1}) = c_{g}(x)c_{g}(y), \]

   so \(c_{g}\) is a homomorphism. Moreover,

   \[ c_{g^{-1}}(c_{g}(x)) = g^{-1}(gxg^{-1})g = x, \]

   and similarly \(c_{g}(c_{g^{-1}}(x)) = x\). Hence \(c_{g^{-1}}\) is the inverse of \(c_{g}\), so \(c_{g}\) is an automorphism.

2. For \(x \in G\),

   \[ (c_{g} \circ c_{h})(x) = g(hxh^{-1})g^{-1} = (gh)x(gh)^{-1} = c_{gh}(x). \]

   Thus \(c_{g} \circ c_{h} = c_{gh}\).

3. Conjugation relabels the entries of a cycle, so

   \[ c_{(1\,2)}((1\,2\,3)) = ((1\,2)(1)\ (1\,2)(2)\ (1\,2)(3)) = (2\,1\,3) = (1\,3\,2). \]

4. If \(G\) is abelian, then \(gx = xg\) for every \(g, x \in G\). Hence

   \[ c_{g}(x) = gxg^{-1} = xgg^{-1} = x. \]

   Thus \(c_{g} = \operatorname{id}_{G}\).

We shall use two elementary properties that are preserved by group isomorphisms.

1. Abelianness is preserved. Suppose that \(\varphi: G \rightarrow H\) is an isomorphism and that \(G\) is abelian. Let \(x, y \in H\). Since \(\varphi\) is surjective, there exist \(a, b \in G\) such that \(x = \varphi(a)\) and \(y = \varphi(b)\). Then

Thus \(H\) is abelian. Applying the same argument to \(\varphi^{-1}\) shows that \(G\) is abelian if and only if \(H\) is abelian.

2. Orders of elements are preserved. Let \(g \in G\). If \(g^{k} = e_{G}\) for some positive integer \(k\), then

Conversely, suppose that \(\varphi(g)^{k} = e_{H}\). Then

Since \(\varphi\) is injective, \(g^{k} = e_{G}\). Thus the same positive integers \(k\) make \(g^{k}\) and \(\varphi(g)^{k}\) equal to the corresponding identity elements. Therefore \(\operatorname{ord}(\varphi(g)) = \operatorname{ord}(g)\).

We now compare the four groups. The group \(D_{4}\) is nonabelian. Indeed, if \(r\) and \(s\) commuted, then \(srs = s^{2}r = r\), whereas the defining relation gives \(srs = r^{-1} = r^{3} \ne r\). The other three groups are abelian. Since an isomorphism preserves abelianness, \(D_{4}\) is not isomorphic to any of the other three groups.

Next, \(C_{8}\) contains an element of order \(8\), namely a generator. In \(C_{4} \times C_{2}\), every element \((x, y)\) satisfies

because \(x^{4} = e\) in \(C_{4}\) while \(y^{2} = e\) and hence \(y^{4} = e\) in \(C_{2}\). Therefore \(C_{4} \times C_{2}\) has no element of order \(8\). Also, every nonidentity element of \(C_{2} \times C_{2} \times C_{2}\) has order \(2\). Since isomorphisms preserve element orders, \(C_{8}\) is not isomorphic to either of these two groups.

Finally, \(C_{4} \times C_{2}\) contains an element of order \(4\). For example, if \(a\) generates \(C_{4}\), then \((a, e)\) has order \(4\). On the other hand, every nonidentity element of \(C_{2} \times C_{2} \times C_{2}\) has order \(2\). Hence these two groups are not isomorphic.

We have distinguished every pair of groups, so the four groups are pairwise nonisomorphic.