Chapter 8: Problem 2
Für \(n=3, \ldots, 7\) gebe man für jeden Primteiler \(p\) von \(n !\) ein Element \(\sigma \in S_{n}\) mit \(o(\sigma)=p\) an.
Short Answer
Expert verified
For each \(n\) from 3 to 7, find permutations corresponding to each prime divisor of \(n!\) using cycles in \(S_n\).
Step by step solution
01
Understand the Problem
Given a number \(n\) ranging from 3 to 7, we must find a permutation \(\sigma\) in the symmetric group \(S_n\) such that the order \(o(\sigma)\) is equal to each prime divisor \(p\) of \(n!\). This means that for each prime \(p\), there should be a permutation whose cycle decomposition leads to an order of \(p\).
02
Calculate Factorials
Calculate the factorial for each \(n\) from 3 to 7 to determine its prime factors. For instance, calculate \(3! = 6\), \(4! = 24\), \(5! = 120\), \(6! = 720\), and \(7! = 5040\). These values will help us identify the prime divisors.
03
Determine Prime Divisors
Identify the prime divisors of each factorial. For example, the prime divisors for \(3! = 6\) are 2 and 3; for \(4! = 24\), they are 2 and 3; for \(5! = 120\), they are 2, 3, and 5; for \(6! = 720\), they are 2, 3, and 5; and for \(7! = 5040\), they are 2, 3, 5, and 7.
04
Find Permutations for Prime Orders
Find permutations in \(S_n\) for each prime divisor's order. A cycle of length \(p\) in \(S_n\) has order \(p\). For instance, the permutation \((1\ 2)\) in \(S_3\) has order 2 and \((1\ 2\ 3)\) has order 3. Continue this process for each \(n\), finding prime-order permutations.
05
List Permutations for Each \(n\)
List the permutations for each \(n\) from 3 to 7:- For \(n = 3\): Prime divisors 2, 3. * \(\sigma = (1\ 2)\) for 2 * \(\sigma = (1\ 2\ 3)\) for 3- For \(n = 4\): Prime divisors 2, 3. * \(\sigma = (1\ 2)\) for 2 * \(\sigma = (1\ 2\ 3)\) for 3- For \(n = 5\): Prime divisors 2, 3, 5. * \(\sigma = (1\ 2)\) for 2 * \(\sigma = (1\ 2\ 3)\) for 3 * \(\sigma = (1\ 2\ 3\ 4\ 5)\) for 5- For \(n = 6\): Prime divisors 2, 3, 5. * \(\sigma = (1\ 2)\) for 2 * \(\sigma = (1\ 2\ 3)\) for 3 * \(\sigma = (1\ 2\ 3\ 4\ 5)\) for 5- For \(n = 7\): Prime divisors 2, 3, 5, 7. * \(\sigma = (1\ 2)\) for 2 * \(\sigma = (1\ 2\ 3)\) for 3 * \(\sigma = (1\ 2\ 3\ 4\ 5)\) for 5 * \(\sigma = (1\ 2\ 3\ 4\ 5\ 6\ 7)\) for 7
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factorials
Factorials are a way of describing the product of all positive integers up to a certain number. They are denoted by an exclamation mark (!). So, for example, 5 factorial is written as 5! and calculated by multiplying 5 x 4 x 3 x 2 x 1, which equals 120. Factorials grow really fast as numbers increase. Here's a quick look at how this works:
- 3! = 3 x 2 x 1 = 6
- 4! = 4 x 3 x 2 x 1 = 24
- 5! = 5 x 4 x 3 x 2 x 1 = 120
- 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
- 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
Prime Divisors
Prime divisors are the prime numbers that divide a given number exactly, leaving no remainder. In the context of factorials, they become quite important. To find these, we break down the factorial number into its smallest prime components.
- For 3!, which is 6, the prime divisors are 2 and 3.
- For 4!, which is 24, the prime divisors are also 2 and 3.
- At 5!, which is 120, they become 2, 3, and 5.
- When we reach 6!, 720, they remain 2, 3, and 5.
- For 7!, at 5040, the prime divisors expand to include 2, 3, 5, and 7.
Permutations
Permutations refer to different ways of arranging a set of items in order. When we talk about permutations in math, we're usually working with a group called a symmetric group, often denoted as \( S_n \), which includes all possible permutations of \( n \) elements.
- For instance, \( S_3 \) has 6 elements because there are 6 ways to order a set of 3 items: 123, 132, 213, 231, 312, 321.
Cycle Decomposition
When analyzing permutations, we often express them in terms of cycles. A cycle in this context refers to a sequence of elements that are permuted among each other. For example, the cycle \((1\, 2\, 3)\) means 1 goes to 2, 2 goes to 3, and 3 goes back to 1. Cycle decomposition breaks down a permutation into these cycles and helps identify the order of a permutation. Understanding Orders: The order of a cycle is the length of the cycle.
- A single swap or 2-cycle, like \((1\, 2)\), has order 2.
- A 3-cycle, such as \((1\, 2\, 3)\), has order 3.