Homework 3: Group Theory

1. Since \(H \le G\) and both groups are finite, Lagrange's theorem gives

   \[[G:H] = \frac{|G|}{|H|} = \frac{84}{12} = 7.\]

2. Since \([G:K]=6\), Lagrange's theorem gives \(|G| = [G:K]|K|\). Therefore

   \[|K| = \frac{84}{6} = 14.\]

3. Since \(L \le H \le G\) and the indices are finite, the multiplicativity of index gives

   \[[G:L] = [G:H][H:L] = 7 \cdot 3 = 21.\]

4. If \(g \in G\) had order 8, then the cyclic subgroup \(\langle g \rangle\) would have order 8. Lagrange's theorem would then imply that \(8 \mid 84\), which is false. Hence \(G\) cannot contain an element of order 8.

1. The identity has order 1. Every nonidentity element \((x, y)\) satisfies \(2(x,y) = (0,0)\), so every nonidentity element has order 2.

2. We have \(4 \mid |V_4|\), but \(V_4\) has no element of order 4.

3. No. The positive divisors of \(|V_4|=4\) are 1, 2, and 4. The trivial subgroup has order 1, each subgroup generated by a nonidentity element has order 2, and \(V_4\) itself has order 4. Thus \(V_4\) has a subgroup of order \(d\) for every divisor \(d\) of 4, so it is not a counterexample to the subgroup statement.

1. Let \(x \in G\) satisfy \(x = x^{-1}\). Multiplying the equality on the right by \(x\) gives \(x^2 = e\). Hence \(\operatorname{ord}(x)\) divides 2. Since \(G\) is finite, Lagrange's theorem also gives \(\operatorname{ord}(x) \mid |G|\). The integer \(|G|\) is odd, so the only positive integer dividing both 2 and \(|G|\) is 1. Therefore \(\operatorname{ord}(x) = 1\) and hence \(x = e\).

2. Consider the map

   \[\iota: G \setminus \{e\} \longrightarrow G \setminus \{e\}, \quad \iota(x) = x^{-1}.\]

   For every \(x \in G \setminus \{e\}\), we have \(\iota(\iota(x)) = x\). By part (a), no element of \(G \setminus \{e\}\) is fixed by \(\iota\). Thus each orbit of \(\iota\) has exactly two elements, namely \(\{x, x^{-1}\}\). These two-element orbits are disjoint and cover \(G \setminus \{e\}\). Hence the nonidentity elements are partitioned into such pairs.

3. Since \(S\) is closed under taking inverses, the inverse map restricts to a map \(S \rightarrow S\). Because \(e \notin S\), part (a) shows that this restriction has no fixed points. As in part (b), \(S\) is therefore a disjoint union of two-element sets \(\{x, x^{-1}\}\). Consequently \(|S|\) is even.

The intersection \(H \cap K\) is a subgroup of both \(H\) and \(K\). By Lagrange's theorem,

Since \(\gcd(9, 14) = 1\), the only positive integer dividing both 9 and 14 is 1. Therefore \(|H \cap K| = 1\), so \(H \cap K = \{e\}\).

1. The residue classes relatively prime to 15 are

   \[U(15) = \{[1], [2], [4], [7], [8], [11], [13], [14]\}.\]

   Hence \(|U(15)| = 8\).

2. We compute powers in \(U(15)\). The identity \([1]\) has order 1. Also

   \[[4]^2 = [16] = [1], \quad [11]^2 = [121] = [1], \quad [14]^2 = [196] = [1],\]

   so \([4]\), \([11]\), and \([14]\) have order 2. Further,

   \[[2]^2 = [4] \ne [1], \quad [2]^4 = [16] = [1],\]

   so \([2]\) has order 4. Likewise,

   \[[7]^2 = [4], \quad [8]^2 = [4], \quad [13]^2 = [4],\]

   and hence \([7]\), \([8]\), and \([13]\) also have order 4.

3. By part (b), every element of \(U(15)\) has order dividing 4. Therefore \(u^4 = [1]\) for every \(u \in U(15)\). Translating this statement into congruence notation gives

   \[a^4 \equiv 1 \pmod{15}\]

   whenever \(\gcd(a, 15) = 1\).

4. Lagrange's theorem uses only the order of the group. Since \(|U(15)| = 8\), it implies \(u^8 = [1]\) for every \(u \in U(15)\). The direct computation in part (b) reveals additional information about the structure of \(U(15)\): no element has order 8, and in fact every element has order dividing 4. That extra information yields the stronger exponent in part (c).

Write \(x = a^k\). Then \(x^6 = e\) exactly when \(40 \mid 6k\), equivalently \(20 \mid 3k\), hence \(20 \mid k\). Modulo 40, \(k \equiv 0, 20\), so the solutions are \(e, a^{20}\). For \(x^5 = e\), we need \(40 \mid 5k\), equivalently \(8 \mid k\). Thus

are the five solutions. Their orders divide 2 and 5, respectively, and therefore divide 40, as required by Lagrange's theorem.

1. \(|D_8| = 16\) and \(|\langle r \rangle| = 8\), so the subgroup has index 2 and is therefore normal.

2. Conjugation by \(r\) fixes every power of \(r\). Also

   \[s r^{2k} s = (srs)^{2k} = r^{-2k} \in \langle r^2 \rangle.\]

   Since \(D_8\) is generated by \(r\) and \(s\), the subgroup is invariant under conjugation by every element of \(D_8\).

3. The quotient has two elements, \(\langle r \rangle\) and \(s\langle r \rangle\). Hence it is cyclic of order 2, so it is isomorphic to \(C_2\).

1. \(N \subseteq Z(Q_8)\), so it is normal.

2. The four cosets are

   \[N, \quad iN = \{i, -i\}, \quad jN = \{j, -j\}, \quad kN = \{k, -k\}.\]

3. Every nonidentity coset has order 2 because, for example, \((iN)^2 = i^2 N = (-1)N = N\). Thus the quotient has four elements and three nonidentity elements of order 2. Therefore

   \[Q_8/N \cong C_2 \times C_2.\]

1. We first verify that \(N\) is a subgroup of \(G\). Since \(1 > 0\), we have \(1 \in N\). If \(x, y \in N\), then \(xy > 0\), so \(xy \in N\). If \(x \in N\), then \(x^{-1} > 0\), so \(x^{-1} \in N\). Hence \(N \le G\). Since \(G = \mathbb{R}^{\times}\) is abelian, every subgroup of \(G\) is normal. Therefore \(N \le G\).

2. Let \(x \in \mathbb{R}^{\times}\). If \(x > 0\), then \(x \in N\), and therefore \(xN = N\). If \(x < 0\), then \(-x > 0\), so \(-x \in N\). Thus

   \[xN = (-1)(-x)N = (-1)N.\]

   Hence every coset is either \(N\) or \((-1)N\). These two cosets are distinct because \(-1 \notin N\). Therefore the quotient has exactly two elements:

   \[G/N = \{N, (-1)N\}.\]

3. Define

   \[\sigma: G/N \longrightarrow C_2, \quad \sigma(xN) = \begin{cases} \overline{0}, & x > 0, \\ \overline{1}, & x < 0. \end{cases}\]

   By part (b), every coset is either \(N\) or \((-1)N\), so this formula assigns one of the two elements of \(C_2\) to each coset. If \(xN = yN\), then \(x^{-1}y \in N\), so \(x^{-1}y > 0\). Hence \(x\) and \(y\) have the same sign, and therefore \(\sigma\) is well defined. The two cosets \(N\) and \((-1)N\) are sent to \(\overline{0}\) and \(\overline{1}\), respectively, so \(\sigma\) is bijective. The product of two nonzero real numbers is positive exactly when the factors have the same sign; this is precisely addition modulo 2 under \(\sigma\). Hence \(\sigma\) is a group isomorphism, and \(G/N \cong C_2\).

   Finally, \(xN = yN \iff x^{-1}y \in \mathbb{R}_{>0} \iff x\) and \(y\) have the same sign. Thus the coset records exactly the sign of the nonzero real number.

1. The set \(N\) contains \((0,0)\). If \((s,s), (t,t) \in N\), then

   \[(s,s) - (t,t) = (s-t, s-t) \in N.\]

   By the subgroup test, \(N \le \mathbb{Z}^2\). Since \(\mathbb{Z}^2\) is abelian, \(N\) is normal in \(G\).

2. Let \((a,b) \in \mathbb{Z}^2\). Since \((b,b) \in N\) we have

   \[(a,b) + N = (a,b) - (b,b) + N = (a-b, 0) + N.\]

   Thus every coset has a representative of the required form.

3. Suppose

   \[(m,0) + N = (n,0) + N.\]

   By the coset equality criterion,

   \[(m,0) - (n,0) = (m-n, 0) \in N.\]

   An element of \(N\) has equal coordinates, so \((m-n, 0) \in N\) implies \(m-n = 0\). Hence \(m = n\). Therefore the representative obtained in part (b) is unique.

4. Define

   \[\Phi: G/N \longrightarrow \mathbb{Z}, \quad \Phi((a,b) + N) = a - b.\]

   Parts (b) and (c) show that each coset has a unique representative \((m, 0)\), and \(\Phi\) sends that coset to \(m\). Hence \(\Phi\) is a bijection. Moreover, for \((a, b), (c,d) \in G\),

   \[\Phi(((a,b)+N) + ((c,d)+N)) = (a+c) - (b+d) = (a-b) + (c-d).\]

   Therefore \(\Phi\) is a group homomorphism. Consequently

   \[\mathbb{Z}^2/N \cong \mathbb{Z}.\]

1. The rotations \(e\) and \(r^2\) commute with both \(r\) and \(s\), while \(r\) and \(r^3\) do not commute with \(s\), and no reflection commutes with \(r\). Hence \(Z(D_4) = \{e, r^2\} = \langle r^2 \rangle\).

2. The cosets are

   \[Z(D_4), \quad rZ(D_4), \quad sZ(D_4), \quad rsZ(D_4).\]

3. Each nonidentity coset has order 2, so the quotient is isomorphic to \(C_2 \times C_2\).

1. Every value \(4a+6b\) is even, so \(\operatorname{im} \varphi \subseteq 2\mathbb{Z}\). Conversely, \(2 = 4(-1) + 6(1)\), so every even integer is in the image. Hence \(\operatorname{im} \varphi = 2\mathbb{Z}\).

2. The equation \(4a+6b = 0\) is equivalent to \(2a+3b = 0\). Thus \(a = 3t\) and \(b = -2t\) for some \(t \in \mathbb{Z}\). Therefore

   \[\operatorname{Ker} \varphi = \langle (3, -2) \rangle.\]

3. By the First Isomorphism Theorem,

   \[\mathbb{Z}^2 / \langle (3, -2) \rangle \cong 2\mathbb{Z} \cong \mathbb{Z}.\]

1. If \([m]_{36} = [n]_{36}\), then \(36 \mid (m-n)\), hence \(12 \mid (m-n)\), so \([m]_{12} = [n]_{12}\). Thus the map is well defined. Every class modulo 12 is the image of the same integer class modulo 36, so the map is surjective.

2. The kernel consists of classes divisible by 12:

   \[\operatorname{Ker} \rho = \{[0]_{36}, [12]_{36}, [24]_{36}\} = \langle [12]_{36} \rangle.\]

3. Therefore

   \[(\mathbb{Z}/36\mathbb{Z})/\langle [12]_{36} \rangle \cong \mathbb{Z}/12\mathbb{Z}.\]

1. Changing \(a\) by a multiple of 12 or \(b\) by a multiple of 18 changes \(a+b\) by a multiple of 6, so the formula is well defined. The element \(([1]_{12}, [0]_{18})\) maps to \([1]_6\), so the map is surjective.

2. \[\operatorname{Ker} \varphi = \{([a]_{12}, [b]_{18}) : a+b \equiv 0 \pmod{6}\}.\]

3. The domain has order \(12 \cdot 18 = 216\) and the image has order 6. By the First Isomorphism Theorem, or equivalently by counting fibers of a surjective homomorphism, the kernel has order \(216/6 = 36\).

4. \[(C_{12} \times C_{18})/\operatorname{Ker} \varphi \cong C_6.\]

1. The kernel is precisely \(SL_n(F)\).

2. The diagonal matrix \(\operatorname{diag}(a, 1, \dots, 1)\) has determinant \(a\), so the determinant map is surjective.

3. Hence

   \[GL_n(F)/SL_n(F) \cong F^{\times}.\]

1. Products of two rotations and of two reflections are rotations, while a product of one rotation and one reflection is a reflection. This matches addition modulo 2, so \(\epsilon(xy) = \epsilon(x) + \epsilon(y)\) for all \(x, y \in D_n\).

2. The kernel is the rotation subgroup \(\langle r \rangle\), and the image is all of \(C_2\) because \(\epsilon(s) = \overline{1}\).

3. The First Isomorphism Theorem gives

   \[D_n/\langle r \rangle \cong C_2.\]

1. \(e^{2\pi it} = 1\) exactly when \(t \in \mathbb{Z}\), so \(\operatorname{Ker} \psi = \mathbb{Z}\).

2. Every point of \(S^1\) has the form \(e^{i\theta} = e^{2\pi i(\theta/2\pi)}\), so \(\psi\) is surjective.

3. Therefore

   \[\mathbb{R}/\mathbb{Z} \cong S^1.\]

1. Since both maps are surjective, the First Isomorphism Theorem gives

   \[G/N \cong H \quad \text{and} \quad G/N \cong K.\]

   Therefore \(H \cong K\).

2. We first check well-definedness. Suppose \(\varphi(g) = \varphi(g')\) for some \(g, g' \in G\). Then

   \[\varphi(g^{-1}g') = \varphi(g)^{-1}\varphi(g') = e_H\]

   so \(g^{-1}g' \in \operatorname{Ker} \varphi = N\). Since \(N = \operatorname{Ker} \psi\), we obtain

   \[\psi(g^{-1}g') = e_K\]

   and hence \(\psi(g) = \psi(g')\). Thus the value \(\Theta(\varphi(g)) = \psi(g)\) does not depend on the choice of \(g\), so \(\Theta\) is well defined.

   Let \(g_1, g_2 \in G\). Then

   \[\Theta(\varphi(g_1)\varphi(g_2)) = \Theta(\varphi(g_1g_2)) = \psi(g_1g_2) = \psi(g_1)\psi(g_2),\]

   which is

   \[\Theta(\varphi(g_1))\Theta(\varphi(g_2)).\]

   Since \(\varphi\) is surjective, this proves that \(\Theta\) is a homomorphism.

   To prove surjectivity, let \(k \in K\). Since \(\psi\) is surjective, there is \(g \in G\) such that \(\psi(g) = k\). Then \(\Theta(\varphi(g)) = k\).

   To prove injectivity, suppose \(\Theta(\varphi(g)) = e_K\). Then \(\psi(g) = e_K\), so \(g \in \operatorname{Ker} \psi = N = \operatorname{Ker} \varphi\). Hence \(\varphi(g) = e_H\). Therefore \(\operatorname{Ker} \Theta = \{e_H\}\) and \(\Theta\) is injective. Thus \(\Theta\) is an isomorphism.

3. No. Take \(G = \mathbb{Z}\), \(H = \mathbb{Z}/5\mathbb{Z}\), and define

   \[\varphi(m) = [m]_5, \quad \psi(m) = [2m]_5.\]

   Both maps are surjective, and both have kernel \(5\mathbb{Z}\). However,

   \[\varphi(1) = [1]_5 \ne [2]_5 = \psi(1).\]

   Thus equal kernels do not force the homomorphisms themselves to be equal.

Since \(a \mid b\), write \(b = qa\) for some positive integer \(q\).

1. Every element of \(b\mathbb{Z}\) has the form \(bk = qak = a(qk)\) for some \(k \in \mathbb{Z}\), and therefore belongs to \(a\mathbb{Z}\). Hence \(b\mathbb{Z} \subseteq a\mathbb{Z}\).

2. We claim that the distinct cosets are

   \[b\mathbb{Z}, \quad a + b\mathbb{Z}, \quad 2a + b\mathbb{Z}, \quad \dots, \quad (q-1)a + b\mathbb{Z}.\]

   Indeed, every element of \(a\mathbb{Z}\) has the form \(ka\). By the division algorithm, write

   \[k = sq + r, \quad 0 \le r < q.\]

   Then

   \[ka = (sq + r)a = sb + ra,\]

   so \(ka + b\mathbb{Z} = ra + b\mathbb{Z}\). Thus every coset occurs in the displayed list.

   To see that these cosets are distinct, suppose

   \[ra + b\mathbb{Z} = sa + b\mathbb{Z}\]

   with \(0 \le r, s < q\). Then \((r-s)a \in b\mathbb{Z}\), so \((r-s)a = tb = tqa\) for some \(t \in \mathbb{Z}\). Since \(a \ne 0\), we have \(r-s = tq\). But \(|r-s| < q\) and hence \(r = s\).

3. There are therefore exactly \(q\) cosets. Since \(q = b/a\),

   \[[a\mathbb{Z} : b\mathbb{Z}] = q = \frac{b}{a}.\]

1. Since \(\mathbb{Z}\) is abelian, \(n\mathbb{Z} \le \mathbb{Z}\). The Second Isomorphism Theorem, applied to \(H = m\mathbb{Z}\) and \(N = n\mathbb{Z}\), gives

   \[\frac{m\mathbb{Z}}{m\mathbb{Z} \cap n\mathbb{Z}} \cong \frac{m\mathbb{Z} + n\mathbb{Z}}{n\mathbb{Z}}.\]

   Using the two subgroup identities supplied in the question, this becomes

   \[\frac{m\mathbb{Z}}{L\mathbb{Z}} \cong \frac{d\mathbb{Z}}{n\mathbb{Z}}.\]

2. Since \(m \mid L\), the preceding exercise gives

   \[\left|\frac{m\mathbb{Z}}{L\mathbb{Z}}\right| = [m\mathbb{Z} : L\mathbb{Z}] = \frac{L}{m}.\]

   Since \(d \mid n\), it also gives

   \[\left|\frac{d\mathbb{Z}}{n\mathbb{Z}}\right| = [d\mathbb{Z} : n\mathbb{Z}] = \frac{n}{d}.\]

3. Isomorphic finite groups have the same order. Therefore

   \[\frac{L}{m} = \frac{n}{d}.\]

   Multiplying by \(md\) yields

   \[dL = mn.\]

   Substituting the definitions of \(d\) and \(L\) gives

   \[\gcd(m,n)\operatorname{lcm}(m,n) = mn.\]

1. Since \(N \le \langle r \rangle\), conjugation by \(r\) preserves \(N\). Also

   \[s r^3 s = (srs)^3 = r^{-3} = r^9 = (r^3)^3 \in N.\]

   Thus conjugation by each generator \(r\) and \(s\) of \(G\) preserves \(N\). It follows that \(N \le G\).

2. The rotation subgroup of \(H\) is \(\langle r^2 \rangle\). Hence

   \[H \cap N = \langle r^2 \rangle \cap \langle r^3 \rangle.\]

   Inside the cyclic group \(\langle r \rangle\) of order 12, this intersection is

   \[\langle r^{\operatorname{lcm}(2,3)} \rangle = \langle r^6 \rangle.\]

   Furthermore, \(HN\) contains \(r^2, r^3\), and \(s\). Since

   \[r = (r^2)^{-1}r^3\]

   we have \(r \in HN\). Thus \(HN\) contains both generators \(r\) and \(s\) of \(G\), and therefore \(HN = G\).

3. Since \(N \le G\), the Second Isomorphism Theorem gives

   \[\frac{H}{H \cap N} \cong \frac{HN}{N}.\]

   Using part (b), this becomes

   \[\frac{H}{\langle r^6 \rangle} \cong \frac{G}{\langle r^3 \rangle}.\]

4. In \(G/N\), the element \(rN\) has order 3, because \(r^3 \in N\) and neither \(r\) nor \(r^2\) belongs to \(N\). The element \(sN\) has order 2, and

   \[(sN)(rN)(sN) = (srs)N = r^{-1}N.\]

   Hence \(G/N\) has the standard dihedral presentation with a rotation of order 3, so

   \[G/N \cong D_3.\]

   By part (c), the quotient \(H/\langle r^6 \rangle\) is isomorphic to the same group. Thus

   \[\frac{H}{\langle r^6 \rangle} \cong \frac{G}{\langle r^3 \rangle} \cong D_3.\]

1. In a cyclic group,

   \[H \cap N = \langle a^{\operatorname{lcm}(4,6)} \rangle = \langle a^{12} \rangle, \quad HN = \langle a^{\gcd(4,6)} \rangle = \langle a^2 \rangle.\]

2. \(|H| = 15\), \(|H \cap N| = 5\), so the first quotient has order 3. Also \(|HN| = 30\) and \(|N| = 10\), so the second quotient has order 3.

3. Since \(N \le G\), the Second Isomorphism Theorem gives

   \[H/(H \cap N) \cong HN/N.\]

4. Both are cyclic groups of order 3, hence are isomorphic to \(C_3\).

1. Since \(N \le G\), the Second Isomorphism Theorem gives

   \[\frac{H}{H \cap N} \cong \frac{HN}{N}.\]

   Using \(H \cap N = \{e\}\) and \(HN = G\), we obtain

   \[H \cong G/N.\]

   In this example \(H \cong C_2\), and hence \(G/N \cong C_2\).

2. The conclusion is not valid. We have

   \[H \times N \cong C_2 \times C_4,\]

   which is abelian. On the other hand, \(D_4\) is nonabelian. Indeed, the relation \(srs = r^{-1}\) implies

   \[sr = r^{-1}s,\]

   and since \(r\) has order 4, \(r^{-1} \ne r\). Hence \(sr \ne rs\). Therefore

   \[G \not\cong H \times N.\]

3. In a direct product, elements coming from the two factors commute with one another. Here the subgroups \(H\) and \(N\) have trivial intersection and satisfy \(HN = G\), but they do not commute elementwise: for example, \(sr \ne rs\). Thus the Second Isomorphism Theorem identifies the quotient \(G/N\) with \(H\), but it does not assert that \(G\) splits as the direct product \(H \times N\).

1. The subgroup is generated by \([12]_{60}\) and has five elements:

   \[[0], [12], [24], [36], [48].\]

2. The Third Isomorphism Theorem gives

   \[(\mathbb{Z}/60\mathbb{Z})/(12\mathbb{Z}/60\mathbb{Z}) \cong \mathbb{Z}/12\mathbb{Z}.\]

3. The quotient has order \(60/5 = 12\), agreeing with \(|\mathbb{Z}/12\mathbb{Z}| = 12\).

1. Consider the homomorphism

   \[\operatorname{sgn} \circ \det: GL_n(\mathbb{R}) \longrightarrow \{\pm 1\}.\]

   Its kernel consists exactly of the matrices with positive determinant. Therefore

   \[GL_n^+(\mathbb{R}) = \operatorname{Ker}(\operatorname{sgn} \circ \det)\]

   is normal in \(GL_n(\mathbb{R})\). Also

   \[SL_n(\mathbb{R}) = \operatorname{Ker}(\det|_{GL_n^+(\mathbb{R})}),\]

   so \(SL_n(\mathbb{R}) \le GL_n^+(\mathbb{R})\).

2. The determinant restricts to a homomorphism

   \[\det: GL_n^+(\mathbb{R}) \longrightarrow \mathbb{R}_{>0}.\]

   Its kernel is \(SL_n(\mathbb{R})\). It is surjective because, for each \(t > 0\), the diagonal matrix

   \[\operatorname{diag}(t, 1, \dots, 1)\]

   has determinant \(t\). By the First Isomorphism Theorem,

   \[GL_n^+(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}_{>0}.\]

3. Apply the Third Isomorphism Theorem to \(SL_n(\mathbb{R}) \le GL_n^+(\mathbb{R}) \le GL_n(\mathbb{R})\). It gives

   \[\frac{GL_n(\mathbb{R})/SL_n(\mathbb{R})}{GL_n^+(\mathbb{R})/SL_n(\mathbb{R})} \cong GL_n(\mathbb{R})/GL_n^+(\mathbb{R}).\]

   The subgroup \(GL_n^+(\mathbb{R})\) has index 2 in \(GL_n(\mathbb{R})\), because an invertible real matrix has either positive or negative determinant and matrices of both types exist. Hence

   \[GL_n(\mathbb{R})/GL_n^+(\mathbb{R}) \cong C_2.\]

4. Under the determinant identification \(GL_n(\mathbb{R})/SL_n(\mathbb{R}) \cong \mathbb{R}^{\times}\), the subgroup \(GL_n^+(\mathbb{R})/SL_n(\mathbb{R})\) corresponds to \(\mathbb{R}_{>0}\). Hence the quotient in part (c) is identified with

   \[\mathbb{R}^{\times}/\mathbb{R}_{>0}.\]

   There are exactly two cosets of \(\mathbb{R}_{>0}\) in \(\mathbb{R}^{\times}\), namely \(\mathbb{R}_{>0}\) and \(-\mathbb{R}_{>0}\). Under the determinant map, these correspond respectively to matrices of positive determinant and matrices of negative determinant. Therefore the quotient has two elements and is isomorphic to \(C_2\). Two matrices determine the same element of this final quotient exactly when their determinants have the same sign. Thus the quotient records exactly the sign of the determinant.

Every subgroup of \(\mathbb{Z}\) has the form \(d\mathbb{Z}\). It contains \(12\mathbb{Z}\) exactly when \(d \mid 12\). Thus the subgroups are

Their images in \(\mathbb{Z}/12\mathbb{Z}\) are respectively

These are all subgroups of the quotient by the Correspondence Theorem.

1. By the Correspondence Theorem, the subgroups \(H\) satisfying \(N \subseteq H \subseteq G\) correspond bijectively to the subgroups of \(G/N\). Since \(|G/N| = 13\) is prime, Lagrange's theorem implies that every subgroup of \(G/N\) has order 1 or 13. Thus \(G/N\) has only the trivial subgroup and itself. The corresponding subgroups of \(G\) are therefore

   \[N \quad \text{and} \quad G.\]

   Hence there are exactly two such subgroups.

2. There is no subgroup \(H\) with \(N \subsetneq H \subsetneq G\). Thus \(N\) is a maximal proper subgroup of \(G\).

3. It cannot happen that \(|G/N| = 12\). To see the more general reason, put \(Q = G/N\) and suppose that \(Q\) is finite and has no nontrivial proper subgroup. Choose \(x \in Q\) with \(x \ne e\). Then \(\langle x \rangle\) is a nontrivial subgroup of \(Q\), so the assumption forces

   \[\langle x \rangle = Q.\]

   Thus \(Q\) is cyclic. If \(Q\) were composite, write \(|Q| = ab\) with \(1 < a < |Q|\). A cyclic group of composite order has a nontrivial proper subgroup; for example, if \(Q = \langle x \rangle\), then \(\langle x^a \rangle\) is nontrivial and proper. This contradicts the assumption on \(Q\). Therefore \(Q\) must be prime.

   Consequently, whenever \(G/N\) is finite, the absence of a subgroup strictly between \(N\) and \(G\) forces \(|G/N|\) to be prime.

1. Under the quotient identification induced by the determinant, the subgroup corresponding to \(\mathbb{R}_{>0}\) is its inverse image under \(\det\). Therefore the corresponding subgroup is

   \[\det^{-1}(\mathbb{R}_{>0}) = \{A \in GL_n(\mathbb{R}) : \det A > 0\} = GL_n^+(\mathbb{R}).\]

2. The subgroup \(\mathbb{R}_{>0}\) has index 2 in \(\mathbb{R}^{\times}\), because the two cosets are the positive and negative real numbers. The Correspondence Theorem preserves indices. Hence

   \[[GL_n(\mathbb{R}) : GL_n^+(\mathbb{R})] = 2.\]

3. The subgroup \(\mathbb{R}_{>0}\) is normal in \(\mathbb{R}^{\times}\) because \(\mathbb{R}^{\times}\) is abelian. By the normality clause of the Correspondence Theorem, its corresponding subgroup \(GL_n^+(\mathbb{R})\) is normal in \(GL_n(\mathbb{R})\).

4. The identity coset is \(GL_n^+(\mathbb{R})\) itself and consists of all matrices with positive determinant. If \(B \in GL_n(\mathbb{R})\) has negative determinant, then the other coset is \(BGL_n^+(\mathbb{R})\).

   For every \(A \in GL_n^+(\mathbb{R})\), we have \(\det(BA) = \det(B)\det(A) < 0\), so this coset consists entirely of matrices with negative determinant. Conversely, if \(C\) has negative determinant, then

   \[\det(B^{-1}C) = \frac{\det C}{\det B} > 0,\]

   so \(B^{-1}C \in GL_n^+(\mathbb{R})\) and therefore \(C \in BGL_n^+(\mathbb{R})\). Thus the two cosets are exactly the positive-determinant and negative-determinant matrices.

1. We have

   \[N = \{(a,b) \in \mathbb{Z}^2 : a+2b \equiv 0 \pmod{6}\}.\]

   The map is surjective because \(\pi(1,0) = [1]_6\) generates \(\mathbb{Z}/6\mathbb{Z}\).

2. By the First Isomorphism Theorem,

   \[\mathbb{Z}^2/N \cong \mathbb{Z}/6\mathbb{Z}.\]

3. The four subgroups of \(\mathbb{Z}/6\mathbb{Z}\) are

   \[\{[0]\}, \quad \langle [3] \rangle = \{[0], [3]\}, \quad \langle [2] \rangle = \{[0], [2], [4]\}, \quad \mathbb{Z}/6\mathbb{Z}.\]

   Their inverse images under \(\pi\) are respectively \(N\),

   \[\begin{aligned}H_2 &= \{(a,b) \in \mathbb{Z}^2 : a+2b \equiv 0 \pmod{3}\}, \\ H_3 &= \{(a,b) \in \mathbb{Z}^2 : a \equiv 0 \pmod{2}\},\end{aligned}\]

   and \(\mathbb{Z}^2\). Here the description of \(H_3\) uses the fact that \(a+2b\) is even exactly when \(a\) is even.

4. Correspondence preserves indices. Thus

   \[[\mathbb{Z}^2 : N] = 6, \quad [\mathbb{Z}^2 : H_2] = 3, \quad [\mathbb{Z}^2 : H_3] = 2, \quad [\mathbb{Z}^2 : \mathbb{Z}^2] = 1.\]

1. The quotient \(C_2 \times C_2\) has five subgroups: the trivial subgroup, its three subgroups of order 2, and the whole group.

2. The corresponding subgroups of \(D_4\) are

   \[\langle r^2 \rangle, \quad \langle r \rangle, \quad \langle r^2, s \rangle, \quad \langle r^2, rs \rangle, \quad D_4.\]

3. Every subgroup of the abelian quotient \(D_4/N\) is normal. By the normality part of the Correspondence Theorem, every subgroup in the displayed list is therefore normal in \(D_4\).