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Given \( \sigma = \left( {135}\right) ,\tau = \left( {27}\right) ,\sigma ,\tau \in {S}_{7} \) ; let us compute \( {\sigma \tau } \) . | Notice right away that every number affected by \( \tau \) is unaffected by \( \sigma \) ; and vice versa. Since the two cycles always remain separate, it is appropriate to represent \( {\sigma \tau } \) as (135)(27), because the cycles don’t reduce any farther. \( \blacklozenge \) | Yes |
Example 11.3.21. Suppose \( \mu \in {S}_{7} \) and \( \mu = \left( \begin{array}{lllllll} 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ 6 & 1 & 4 & 3 & 7 & 2 & 5 \end{array}\right) \) . | Then\n\n- \( 1 \rightarrow 6,6 \rightarrow 2 \), and \( 2 \rightarrow 1 \) ; therefore we have the cycle (162).\n\n- \( 3 \rightarrow 4 \) and \( 4 \rightarrow 3 \) ; therefore we have (34).\n\n- Finally, \( 5 \rightarrow 7 \) and \( 7 \rightarrow 5 \) ; therefore we have (57).\n\nHence \( \mu = \left( {162}\right) \le... | Yes |
Proposition 11.3.27. Disjoint cycles commute: that is, given two disjoint cycles \( \sigma = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right) \) and \( \tau = \left( {{b}_{1},{b}_{2},\ldots ,{b}_{k}}\right) \) we have\n\n\[ \n{\sigma \tau } = {\tau \sigma } = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right) \left( {{b}_{... | Proof. We present this proof as a fill-in-the-blanks exercise:\n\nExercise 11.3.28. Fill in the blanks to complete the proof:\n\nRecall that permutations are defined as bijections on a set \( X \) . In order to show that the two permutations \( {\sigma \tau } \) and \( {\tau \sigma } \) are equal, it’s enough to show t... | No |
Given the permutations \( \mu = \left( {257}\right) \left( {134}\right) \) and \( \rho = \) (265)(137) in \( {S}_{7} \), write \( {\mu \rho } \) in cycle notation. | \[ \text{-}1 \rightarrow 3,3 \rightarrow 3,3 \rightarrow 4\text{, and}4 \rightarrow 4\text{; therefore}1 \rightarrow 4\text{.} \]\n\[ \text{-}4 \rightarrow 4,4 \rightarrow 4,4 \rightarrow 1\text{, and}1 \rightarrow 1\text{; therefore}4 \rightarrow 1\text{.} \]\nThis gives us the cycle (14). Continuing,\n\[ \text{-}2 \r... | Yes |
Example 11.3.32. Find the product (156)(2365)(123) in \( {S}_{6} \) . | \[ \text{-}1 \rightarrow 2,2 \rightarrow 3\text{, and}3 \rightarrow 3\text{; therefore}1 \rightarrow 3\text{.} \]\n\n- \( 3 \rightarrow 1,1 \rightarrow 1 \), and \( 1 \rightarrow 5 \) ; therefore \( 3 \rightarrow 5 \) .\n\n\[ \text{-}5 \rightarrow 5,5 \rightarrow 2\text{, and}2 \rightarrow 2\text{; therefore}5 \rightar... | Yes |
(a) Every permutation \( \sigma \) in \( {S}_{n} \) can be written either as the identity, a single cycle, or as the product of disjoint cycles. | Proof. Let’s begin with (a) We can assume that \( X = \{ 1,2,\ldots, n\} \) . Let \( \sigma \in {S}_{n} \), and define \( {X}_{1} = \left\{ {1,\sigma \left( 1\right) ,{\sigma }^{2}\left( 1\right) ,\ldots }\right\} \) . The set \( {X}_{1} \) is finite since \( X \) is finite. Therefore the sequence \( 1,\sigma \left( 1\... | Yes |
We know that every permutation in \( {S}_{5} \) is the product of disjoint cycles. Let us list all possible cycle lengths and number of cycles for the permutations of \( {S}_{5} \) . | - First of all, \( {S}_{5} \) contains the identity, which has no cycles.\n\n- Second, some permutations in \( {S}_{5} \) consist of a single cycle. The single cycle could have length \( 2,3,4 \), or 5 (remember, we don’t count cycles of length 1).\n\n- Third, some permutations in \( {S}_{5} \) consist of the product o... | Yes |
Consider the product (1264)(1264), which we may also write as \( {\left( {1264}\right) }^{2} \). | (1) Notice for all elements \( x \neq 1,2,6,4, x \) stays put in (1264); hence \( x \) stays put in \( {\left( {1264}\right) }^{2} \) . So the product \( {\left( {1264}\right) }^{2} \) does not involve any elements except \( 1,2,6 \) and 4 .\n\n(2) Now let’s look at what happens when \( x = 1,2,6 \), or 4 . By squaring... | Yes |
Proposition 11.4.6. The order of a cycle is always equal to the cycle's length. | Proof. To prove this, we essentially have to prove two things:\n\n(A) If \( \sigma \) is a cycle of length \( k \), then \( {\sigma }^{k} = \mathrm{{id}} \) ;\n\n(B) If \( \sigma \) is a cycle of length \( k \), then \( {\sigma }^{j} \neq \operatorname{id}\forall j : 1 \leq j < k \) .\n\nThe proof for (A) follows the s... | No |
Example 11.4.9. Here's a nice application of Proposition 11.4.6, which simply uses rules of function composition. | \[ {\left( {1264}\right) }^{6} = {\left( {1264}\right) }^{4}{\left( {1264}\right) }^{2} = \mathsf{{id}}\;\left( {16}\right) \left( {24}\right) = \left( {16}\right) \left( {24}\right) \] | Yes |
Example 11.4.13. Let \( \tau = \left( {24}\right) \left( {16}\right) \) . Notice that (24) and (16) are disjoint, so they commute (recall Proposition 11.3.27). We also know that permutations are associative under composition. So we may compute \( {\tau }^{2} \) as follows: | \[ {\tau }^{2} = \left( {\left( {24}\right) \left( {16}\right) }\right) \left( {\left( {24}\right) \left( {16}\right) }\right) \]\n\[ = \left( {24}\right) \left( {\left( {16}\right) \left( {24}\right) }\right) \left( {16}\right) \;\text{(associative)} \]\n\[ = \left( {24}\right) \left( {\left( {24}\right) \left( {16}\r... | Yes |
Proposition 11.4.17. If \( \sigma \) and \( \tau \) are disjoint cycles, then\n\n\[ \left| {\sigma \tau }\right| = \operatorname{lcm}\left( {\left| \sigma \right| ,\left| \tau \right| }\right) \]\n\nwhere 'lcm' denotes least common multiple. | Proof. Let \( j \equiv \left| \sigma \right|, k \equiv \left| \tau \right| \), and \( m \equiv \operatorname{lcm}\left( {k, j}\right) \) . Then it’s enough to prove:\n\n(i) \( {\left( \sigma \tau \right) }^{m} = \mathrm{{id}} \) ;\n\n(ii) \( {\left( \sigma \tau \right) }^{n} \neq \) id if \( n \in \mathbb{N} \) and \( ... | Yes |
Proposition 11.4.24. Every cycle can be written as the product of transpositions: | Proof. The proof involves checking that left and right sides of the equation agree when they act on any \( {a}_{j} \) . We know that the cycle acting on \( {a}_{j} \) gives \( {a}_{j + 1} \) (or \( {a}_{1} \), if \( j = n \) ); while the product of transpositions sends \( {a}_{j} \) first to \( {a}_{1} \), then to \( {... | No |
Proposition 11.4.25. Any permutation of a finite set containing at least two elements can be written as the product of transpositions. | Proof. First write the permutation as a product of cycles: then write each cycle as a product of transpositions. | No |
Proposition 11.4.29. Suppose \( \mu \) is a cycle: \( \mu = \left( {{a}_{1}{a}_{2}\ldots {a}_{n}}\right) \) . Then \( {\mu }^{-1} = \left( {{a}_{1}{a}_{n}{a}_{n - 1}\ldots {a}_{2}}\right) . \) | Proof. By Proposition 11.4.24 we can write\n\n\[ \mu = \left( {{a}_{1}{a}_{n}}\right) \left( {{a}_{1}{a}_{n - 1}}\right) \cdots \left( {{a}_{1}{a}_{3}}\right) \left( {{a}_{1}{a}_{2}}\right) . \]\n\nNow consider first just the last two transpositions in this expression. In the Functions chapter, we proved the formula \(... | Yes |
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