Let's break down each sub-question in **two steps** each, with explanations. --- ### **6.1: Write \( p \) as a function** **Step 1:** Read the permutation table: \[ p = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 2 & 3 & 4 & 1 & 5 & 8 & 6 & 7 \end{pmatrix} \] **Step 2:** Express as function mappings: - \(p(1) = 2\), \(p(2) = 3\), \(p(3) = 4\), \(p(4) = 1\), \(p(5) = 5\), \(p(6) = 8\), \(p(7) = 6\), \(p(8) = 7\). **Final answer:** \[ p(1) = 2,\, p(2) = 3,\, p(3) = 4,\, p(4) = 1,\, p(5) = 5,\, p(6) = 8,\, p(7) = 6,\, p(8) = 7 \] --- ### **6.2: Write \( p \) as a product of disjoint cycles** **Step 1:** Start at 1 and follow the mapping until it returns to 1: 1 → 2 → 3 → 4 → 1, gives cycle (1 2 3 4). Continue with the next unused number (5): - 5 → 5, so (5). Continue with 6: 6 → 8 → 7 → 6, gives (6 8 7). **Step 2:** Write out all cycles (omit (5) since it's the identity): **Final answer:** \[ p = (1\,2\,3\,4)(6\,8\,7) \] or with all cycles: \((1\,2\,3\,4)(5)(6\,8\,7)\) --- ### **6.3: Write \( p \) as a product of transpositions** **Step 1:** Recall: \((a\,b\,c\,d) = (a\,d)(a\,c)(a\,b)\), \((a\,b\,c) = (a\,c)(a\,b)\). So, - \((1\,2\,3\,4) = (1\,4)(1\,3)(1\,2)\) - \((6\,8\,7) = (6\,7)(6\,8)\) **Step 2:** Write the product: \[ p = (1\,4)(1\,3)(1\,2)(6\,7)(6\,8) \] **Final answer:** \[ p = (1\,4)(1\,3)(1\,2)(6\,7)(6\,8) \] --- ### **6.4: Is \( p \) even or odd?** **Step 1:** Count the number of transpositions in the product above: - (1 4), (1 3), (1 2): 3 transpositions - (6 7), (6 8): 2 transpositions - Total: 5 transpositions **Step 2:** If the total is **odd**, \(p\) is **odd**; if **even**, \(p\) is **even**. **Final answer:** \[ p \text{ is an odd permutation.} \] --- Would you like me to continue with parts 6.5–6.8?

Certainly! Let's continue step-by-step for parts 6.5 to 6.8. --- ### **6.5: Compute \( p^2 \)** **Step 1:** Recall \( p \) as the permutation: \[ p = (1\,2\,3\,4)(6\,8\,7) \] from earlier. **Step 2:** Square each cycle: - For a 4-cycle \((a\,b\,c\,d)\), \((a\,b\,c\,d)^2 = (a\,c)(b\,d)\). - For a 3-cycle \((x\,y\,z)\), \((x\,y\,z)^2 = (x\,z\,y)\). Applying: - \((1\,2\,3\,4)^2 = (1\,3)(2\,4)\) - \((6\,8\,7)^2 = (6\,7\,8)\) **Final answer:** \[ p^2 = (1\,3)(2\,4)(6\,7\,8) \] --- ### **6.6: Compute \( p^{-1} \)** **Step 1:** The inverse of a cycle is its inverse cycle: - \((a\,b\,c\,d)^{-1} = (a\,d\,c\,b)\) - \((x\,y\,z)^{-1} = (z\,y\,x)\) So: - \((1\,2\,3\,4)^{-1} = (1\,4\,3\,2)\) - \((6\,8\,7)^{-1} = (7\,8\,6)\) **Step 2:** Write \( p^{-1} \) as the product of these cycles: \[ p^{-1} = (1\,4\,3\,2)(7\,8\,6) \] --- ### **6.7: Determine the period \( k \) such that \( p^k = e \) (identity)** **Step 1:** The period is the least common multiple (LCM) of the cycle lengths: - Cycle lengths are 4 and 3. - \(\text{LCM}(4, 3) = 12\). **Step 2:** Therefore, the smallest positive integer \(k\) with \(p^k = e\) is 12. **Final answer:** \[ \boxed{k=12} \] --- ### **6.8: Find a permutation \( g \) such that \( g \circ p = q \)**, where \[ q = (1\,3\,4\,6\,5\,8\,7\,2) \] **Step 1:** Since \(g \circ p = q\), then: \[ g = q \circ p^{-1} \] We already have: - \( p^{-1} = (1\,4\,3\,2)(7\,8\,6) \), - \( q = (1\,3\,4\,6\,5\,8\,7\,2) \). **Step 2:** Compute \( g = q \circ p^{-1} \). Applying \( p^{-1} \) first, then \( q \): - Find \( g(i) = q(p^{-1}(i)) \) for each \( i \in \{1,\dots,8\} \). Calculations: - \( p^{-1}(1) = 4 \), \( g(1) = q(4) = 6 \) - \( p^{-1}(2) = 1 \), \( g(2) = q(1) = 3 \) - \( p^{-1}(3) = 2 \), \( g(3) = q(2) = 2 \) - \( p^{-1}(4) = 3 \), \( g(4) = q(3) = 4 \) - \( p^{-1}(5) = 5 \), \( g(5) = q(5) = 8 \) - \( p^{-1}(6) = 8 \), \( g(6) = q(8) = 7 \) - \( p^{-1}(7) = 6 \), \( g(7) = q(6) = 5 \) - \( p^{-1}(8) = 7 \), \( g(8) = q(7) = 1 \) Thus, \( g \) as a permutation: \[ g = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 6 & 3 & 2 & 4 & 8 & 7 & 5 & 1 \end{pmatrix} \] --- **Summary:** - **6.1:** \( p(1)=2, p(2)=3, p(3)=4, p(4)=1, p(5)=5, p(6)=8, p(7)=6, p(8)=7 \) - **6.2:** \( p = (1\,2\,3\,4)(6\,8\,7) \) - **6.3:** \( p = (1\,4)(1\,3)(1\,2)(6\,7)(6\,8) \) - **6.4:** \( p \) is an **odd** permutation. - **6.5:** \( p^2 = (1\,3)(2\,4)(6\,7\,8) \) - **6.6:** \( p^{-1} = (1\,4\,3\,2)(7\,8\,6) \) - **6.7:** The period \(k=12\). - **6.8:** \( g = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 6 & 3 & 2 & 4 & 8 & 7 & 5 & 1 \end{pmatrix} \) Let me know if you'd like further clarification!

Certainly! Here's a comprehensive step-by-step solution for all parts (6.1 to 6.8): --- ### **6.1: Write \( p \) as a function** **Step 1:** Given the permutation table: \[ p = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 2 & 3 & 4 & 1 & 5 & 8 & 6 & 7 \end{pmatrix} \] **Step 2:** Map each element: \[ p(1) = 2,\quad p(2) = 3,\quad p(3) = 4,\quad p(4) = 1,\quad p(5) = 5,\quad p(6) = 8,\quad p(7) = 6,\quad p(8) = 7 \] **Final answer:** \[ \boxed{ p(1)=2,\quad p(2)=3,\quad p(3)=4,\quad p(4)=1,\quad p(5)=5,\quad p(6)=8,\quad p(7)=6,\quad p(8)=7 } \] --- ### **6.2: Write \( p \) as a product of disjoint cycles** **Step 1:** Start with 1: \[ 1 \to p(1)=2,\quad 2 \to p(2)=3,\quad 3 \to p(3)=4,\quad 4 \to p(4)=1 \] Cycle: \((1\,2\,3\,4)\). Next, 5: \[ 5 \to p(5)=5 \] Cycle: \((5)\) (identity, can be omitted). Next, 6: \[ 6 \to p(6)=8,\quad 8 \to p(8)=7,\quad 7 \to p(7)=6 \] Cycle: \((6\,8\,7)\). **Step 2:** Write the permutation as a product of disjoint cycles: \[ \boxed{ p = (1\,2\,3\,4)(6\,8\,7) } \] --- ### **6.3: Write \( p \) as a product of transpositions** **Step 1:** Break down cycles: - \((1\,2\,3\,4) = (1\,4)(1\,3)(1\,2)\) - \((6\,8\,7) = (6\,7)(6\,8)\) **Step 2:** Final product: \[ \boxed{ p = (1\,4)(1\,3)(1\,2)(6\,7)(6\,8) } \] --- ### **6.4: Is \( p \) even or odd?** **Step 1:** Count transpositions: - From above: 3 transpositions for the 4-cycle, 2 for the 3-cycle. - Total transpositions: \(3 + 2 = 5\). **Step 2:** Since 5 is odd, \( p \) is an **odd permutation**. \[ \boxed{ p \text{ is odd} } \] --- ### **6.5: Compute \( p^2 \)** **Step 1:** Square the cycles: - \((a\,b\,c\,d)^2 = (a\,c)(b\,d)\) - \((x\,y\,z)^2 = (x\,z\,y)\) Thus: \[ p^2 = (1\,2\,3\,4)^2 \cdots \] \[ (1\,2\,3\,4)^2 = (1\,3)(2\,4),\quad (6\,8\,7)^2 = (6\,7\,8) \] **Step 2:** Final result: \[ \boxed{ p^2 = (1\,3)(2\,4)(6\,7\,8) } \] --- ### **6.6: Compute \( p^{-1} \)** **Step 1:** Inverse of cycles: - \((a\,b\,c\,d)^{-1} = (a\,d\,c\,b)\) - \((x\,y\,z)^{-1} = (z\,y\,x)\) So: \[ p^{-1} = (1\,4\,3\,2)(7\,8\,6) \] **Step 2:** Write explicitly: \[ \boxed{ p^{-1} = (1\,4\,3\,2)(7\,8\,6) } \] --- ### **6.7: Determine the period \( k \) such that \( p^k = e \)** **Step 1:** Cycle lengths are 4 and 3. **Step 2:** The order of permutation is the least common multiple: \[ \mathrm{LCM}(4,3) = 12 \] \[ \boxed{ \text{Period } k=12 } \] --- ### **6.8: Find permutation \( g \) such that \( g \circ p = q \), where** \[ q = (1\,3\,4\,6\,5\,8\,7\,2) \] **Step 1:** Recall that: \[ g = q \circ p^{-1} \] Using \( p^{-1} = (1\,4\,3\,2)(7\,8\,6) \), compute \( g \): \[ g(i) = q(p^{-1}(i)) \] Calculate for each \( i \): | \(i\) | \(p^{-1}(i)\) | \(g(i) = q(p^{-1}(i))\) | |--------|--------------|------------------------| | 1 | 4 | \(q(4) = 6\) | | 2 | 1 | \(q(1) = 3\) | | 3 | 2 | \(q(2) = 2\) | | 4 | 3 | \(q(3) = 4\) | | 5 | 5 | \(q(5) = 8\) | | 6 | 8 | \(q(8) = 7\) | | 7 | 6 | \(q(6) = 5\) | | 8 | 7 | \(q(7) = 1\) | **Step 2:** Write \( g \) as a permutation: \[ g = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 6 & 3 & 2 & 4 & 8 & 7 & 5 & 1 \end{pmatrix} \] --- **In summary:** | Part | Result | |---------|-------------------------------------------------------------| | 6.1 | \( p(1)=2,\ p(2)=3,\ p(3)=4,\ p(4)=1,\ p(5)=5,\ p(6)=8,\ p(7)=6,\ p(8)=7 \) | | 6.2 | \( p = (1\,2\,3\,4)(6\,8\,7) \) | | 6.3 | \( p = (1\,4)(1\,3)(1\,2)(6\,7)(6\,8) \) | | 6.4 | \( p \) is **odd** | | 6.5 | \( p^2 = (1\,3)(2\,4)(6\,7\,8) \) | | 6.6 | \( p^{-1} = (1\,4\,3\,2)(7\,8\,6) \) | | 6.7 | Period \(k=12\) | | 6.8 | \( g = \begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 6 & 3 & 2 & 4 & 8 & 7 & 5 & 1 \end{pmatrix} \) | Let me know if you'd like further assistance!