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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 \( \mu = \left( {162}\right) \left( {34}\right) \left( {57}\right) \), as we may verify by computing the product \( \left( {162}\right) \circ \) (34) \( \circ \) (57) directly. | 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\[ \n{\sigma \tau } = {\tau \sigma } = \left( {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right) \left( {{b}_{1},{b}_{2},\ldots ,{b}_{k}}\right) \n\] | 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 that they are the same function. In other words, we just need to show that \( {\sigma \tau }\left( x\right) = < 1 > \) for all \( x \in X \) .\n\nWe’ll define \( A = \left\{ {{a}_{1},{a}_{2},\ldots ,{a}_{j}}\right\} \) and \( B = \left\{ {{b}_{1},{b}_{2},\ldots ,{b}_{k}}\right\} \) . By hypothesis \( A \) and \( B \) are disjoint, so \( A < 2 > B = < 3 > \) . Given an arbitrary \( x \in X \) , there are three possibilities: (i) \( x \in A \) and \( x \notin B \) ; (ii) \( x \in < 4 > \) and \( x \notin \underline{ < 6 > } \) ; (iii) \( x \notin \underline{ < 7 > } \) and \( x \notin \underline{ < 8 > } \) .\n\n(i) In this case, since \( x \notin B \) it follows that \( \tau \left( x\right) = x \) . We then have \( {\sigma \tau }\left( x\right) = \sigma \left( {\tau \left( x\right) }\right) = \sigma \left( x\right) \) . Furthermore, since \( x \in A \) it follows that \( \sigma \left( x\right) \in A \), so \( \sigma \left( x\right) \notin B \) . We then have \( {\tau \sigma }\left( x\right) = \tau \left( {\sigma \left( x\right) }\right) = \sigma \left( x\right) \) . It follows that \( {\sigma \tau }\left( x\right) = {\tau \sigma }\left( x\right) \).\n\n(ii) In this case, since \( x \notin < 9 > \) it follows that \( < {10} > \left( x\right) = x \) . We then have \( {\tau \sigma }\left( x\right) = < {11} > = < {12} > \left( x\right) \) . Furthermore, since \( x \in \) < 13 > it follows that \( - {14} > \left( x\right) \in \underline{ < {15} > } \), so \( - {16} > \left( x\right) \notin \) \( < {17} > \) . We then have \( {\sigma \tau }\left( x\right) = < {18} > = < {19} > \left( x\right) \) . It follows that \( {\sigma \tau }\left( x\right) = {\tau \sigma }\left( x\right) \).\n\n(iii) In this case, since \( x \notin A \) it follows that \( < {20} > \left( x\right) = x \) . Similarly since \( x \notin < {21} > \) it follows that \( < {22} > \left( x\right) = x \) .We then have \( {\tau \sigma }\left( x\right) = \underline{ < {23} > } \) and \( {\sigma \tau }\left( x\right) = \underline{ < {24} > } \) . It follows that \( {\sigma \tau }\left( x\right) = \) \( {\tau \sigma }\left( x\right) \).\n\nIn all three cases we have \( {\sigma \tau }\left( x\right) = < {25} > \), so therefore \( {\sigma \tau } = {\tau \sigma } \) . \( \diamond \) | 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\[ \text{-}2 \rightarrow 2,2 \rightarrow 6,6 \rightarrow 6\text{, and}6 \rightarrow 6\text{; therefore}2 \rightarrow 6\text{.} \]\n\[ \text{-}6 \rightarrow 6,6 \rightarrow 5,5 \rightarrow 5\text{, and}5 \rightarrow 7\text{; therefore}6 \rightarrow 7\text{.} \]\n\[ \text{-}7 \rightarrow 1,1 \rightarrow 1,1 \rightarrow 3\text{, and}3 \rightarrow 3\text{; therefore}7 \rightarrow 3\text{.} \]\n\[ \text{-}3 \rightarrow 7,7 \rightarrow 7,7 \rightarrow 7\text{, and}7 \rightarrow 2\text{; therefore}3 \rightarrow 2\text{.} \]\nSo we have the cycle (2673). Now the only input not included in our cycles is 5 , so logically it should stay put. But let's test it just in case we made a mistake in our work above.\n- \( 5 \rightarrow 5,5 \rightarrow 2,2 \rightarrow 2 \), and \( 2 \rightarrow 5 \) ; therefore 5 does indeed stay put.\nSo, we finally have: \( {\mu \rho } = \left( {14}\right) \left( {2673}\right) \) | 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\[ \text{-}5 \rightarrow 5,5 \rightarrow 2\text{, and}2 \rightarrow 2\text{; therefore}5 \rightarrow 2\text{.} \]\n\n\[ \text{-}2 \rightarrow 3,3 \rightarrow 6\text{, and}6 \rightarrow 1\text{; therefore}2 \rightarrow 1\text{.} \]\n\nSo we have (1352).\n\n- 4 does not appear in any of the cycles, so we know it won't be acted on by any of the cycles. Hence 4 stays put.\n\n- \( 6 \rightarrow 6,6 \rightarrow 5 \), and \( 5 \rightarrow 6 \) ; hence 6 stays put.\n\nTherefore (156)(2365)(123) \( = \left( {1352}\right) \) . | 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 \( 1,\sigma \left( 1\right) ,{\sigma }^{2}\left( 1\right) ,\ldots \) must repeat. Let \( {j}_{1} \) be the first index where the sequence repeats, so that \( {\sigma }^{{j}_{1}}\left( 1\right) = {\sigma }^{k}\left( 1\right) \) for some \( k < {j}_{1} \) . Then if we apply \( {\sigma }^{-1} \) to both sides of the equation we get \( {\sigma }^{{j}_{1} - 1}\left( 1\right) = {\sigma }^{k - 1}\left( 1\right) \) . Repeating this \( k - 1 \) more times gives \( {\sigma }^{{j}_{1} - k}\left( 1\right) = 1 \) . This implies that the sequence repeats at index \( {j}_{1} - k \) : but we’ve already specified that \( {j}_{1} \) is the first index where the sequence repeats. The only way this can happen is if \( k = 0 \) . It follows that \( {X}_{1} = \left\{ {1,\sigma \left( 1\right) ,{\sigma }^{2}\left( 1\right) ,\ldots {\sigma }^{{j}_{1} - 1}\left( 1\right) }\right\} \) , where \( {\sigma }^{{j}_{1}}\left( 1\right) = 1 \) .\n\nNow there are two possible cases:\n\n(i) \( {X}_{1} \) accounts for all the integers in \( X \) ; i.e. \( {X}_{1} = X \)\n\n(ii) there are some integers in \( X \) not accounted for in \( {X}_{1} \) (that is, \( X \smallsetminus {X}_{1} \neq \) (1).\n\nIf case (ii) holds, then let \( i \) be the smallest integer in \( X \smallsetminus {X}_{1} \) and define \( {X}_{2} \) by \( \left\{ {i,\sigma \left( i\right) ,{\sigma }^{2}\left( i\right) ,\ldots }\right\} \) . Just as with \( {X}_{1} \), we may conclude that \( {X}_{2} \) is a finite set, and that \( {X}_{2} = \left\{ {i,\sigma \left( i\right) ,\ldots ,{\sigma }^{{j}_{2} - 1}\left( i\right) }\right\} \) where \( {\sigma }^{{j}_{2}}\left( i\right) = i \) .\n\nWe claim furthermore that \( {X}_{1} \) and \( {X}_{2} \) are disjoint. We can see this by contradiction: suppose on the other hand that \( {X}_{1} \) and \( {X}_{2} \) are not disjoint.\n\nThen it must be the case that \( {\sigma }^{p}\left( 1\right) = {\sigma }^{q}\left( i\right) \) for some natural numbers \( p, q \) with \( 0 \leq p < {j}_{1} \) and \( 0 \leq q < {j}_{2} \) . Applying \( \sigma \) to both sides of this equation, gives \( {\sigma }^{p + 1}\left( 1\right) = {\sigma }^{q + 1}\left( i\right) \) . If we continue applying \( \sigma \) to both sides a total of \( {j}_{2} - q \) times then we obtain \( {\sigma }^{p + {j}_{2} - q}\left( 1\right) = {\sigma }^{{j}_{2}}\left( i\right) \) . But since \( {\sigma }^{{j}_{2}}\left( i\right) = \) \( i \), it follows that \( {\sigma }^{p + {j}_{2} - q}\left( 1\right) = i \), which implies that \( i \in {X}_{1} \) . This is a contradiction, because we know \( i \in X \smallsetminus {X}_{1} \) . The contradiction shows that the suppo | 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 of two disjoint cycles. To enumerate these, suppose first that one of the cycles is a cycle of length 2 . Then the other cycle could be a cycle of length 2 (for instance in the case (12)(34)) or a cycle of length 3 (as in the case (14)(235)). There are no other possibilities, because we only have 5 elements to permute, and a larger disjoint cycle would require more elements.\n\n- It's not possible to have three or more disjoint cycles, because that would require at least six elements.\n\nTo summarize then, the possible cycle structures for permutations in \( {S}_{5} \) are:\n\n- The identity\n\n- single cycles of lengths \( 5,4,3 \), or 2\n\n- two disjoint cycles of lengths 2 and 3 ; and two disjoint cycles of lengths 2 and 2 | 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 the cycle, we are applying it twice to each input; hence each input is moved two spots around the wheel (see Figure 11.4.2) . In other words,\n\n\[ 1 \rightarrow 6;\;6 \rightarrow 1;\;2 \rightarrow 4;\;4 \rightarrow 2, \]\n\nAltogether: \( {\left( {1264}\right) }^{2} = \left( {16}\right) \left( {24}\right) \) . | 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 same lines as our investigations in Exercise 11.4.4. In that exercise, we considered separately the elements of \( X \) that are moved by the cycle, and those elements that are not moved by the cycle.\n\nExercise 11.4.7. Prove part (A) by filling in the blanks.\n\nLet \( \sigma \in {S}_{X} \) be an arbitrary cycle of length \( k \) . Then \( \sigma \) can be written as \( \left( {{a}_{1}{a}_{2}\ldots {a}_{k}}\right) \), for some set of elements \( {a}_{1},{a}_{2},\ldots {a}_{k} \) in \( X \) . In order to show that \( {\sigma }^{k} = \mathrm{{id}} \), it is sufficient to show that \( {\sigma }^{k}\left( x\right) = < 1 > \forall x \in X \) . Let \( A \) be the set \( \left\{ {{a}_{1},{a}_{2},\ldots {a}_{k}}\right\} \) . Now for any \( x \in X \), there are two possibilities:\n\n(i) \( x \in X \smallsetminus A \) ;\n\n\( {}^{3} \) This is in keeping with our practice of using \( \left| \ldots \right| \) to denote the \ | 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\[ = \left( {24}\right) \left( {\left( {24}\right) \left( {16}\right) }\right) \left( {16}\right) \;\text{(commutative)} \]\n\[ = \left( {\left( {24}\right) \left( {24}\right) }\right) \left( {\left( {16}\right) \left( {16}\right) }\right) \]\n\[ \text{(associative)} \]\n\[ = \text{id id}\;\left( {2\text{-cycles have order }2}\right) \]\n\[ = \text{id} \] | 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 \( n < m \) .\n\nTo prove (i), first note that \( k \) divides \( m \), so that \( m = j \cdot p \) for some natural number \( p \) . Similarly, \( m = k \cdot q \) for some \( q \in \mathbb{N} \) . It follows:\n\n\[ {\left( \sigma \tau \right) }^{m} = {\sigma }^{m}{\tau }^{m}\;\text{(by Exercise 11.4.15)} \]\n\n\[ = {\sigma }^{j \cdot p}{\tau }^{k \cdot q}\;\text{(by definition of lcm)} \]\n\n\[ = {\left( {\sigma }^{j}\right) }^{p}{\left( {\tau }^{k}\right) }^{q}\;\text{ (by exponentiation rules) }{}^{4} \]\n\n\[ = {\mathrm{{id}}}^{p}{\mathrm{{id}}}^{q}\;\text{ (by definition of order) } \]\n\n\[ = \mathrm{{id}}\;\text{(by definition of}\mathrm{{id}}\text{).} \]\n\nTo prove (ii), let \( n < m \) . It follows either \( k \) or \( j \) does not divide \( n \) . Let’s suppose it’s \( k \) (the case where it’s \( j \) is virtually identical). In this case we must have \( n = p \cdot k + r \) where \( p,\mathrm{r} \in \mathbb{N} \) and \( r < k \) . It follows:\n\n\[ {\left( \sigma \tau \right) }^{n} = {\sigma }^{n}{\tau }^{n}\;\text{ (by Exercise 11.4.15) } \]\n\n\[ = {\sigma }^{j \cdot p + r}{\tau }^{n}\;\text{ (substitution) } \]\n\n\[ = {\left( {\sigma }^{j}\right) }^{p}{\sigma }^{r}{\tau }^{n}\;\text{ (by exponentiation rules) } \]\n\n\[ = {\operatorname{id}}^{p}{\sigma }^{r}{\tau }^{n}\;\text{ (by definition of order) } \]\n\n\[ = {\sigma }^{r}{\tau }^{n}\;\text{(by definition of identity)} \]\n\nNow since \( r < k \), and \( \left| \sigma \right| = k \), it follows that \( {\sigma }^{r} \neq \) id. Thus there is some \( x \) such that \( {\sigma }^{r}\left( x\right) \neq x \) . But since \( \sigma \) and \( \tau \) are disjoint, it must be the case that \( \tau \left( x\right) = x \) . It follows that:\n\n\[ {\sigma }^{r}{\tau }^{n}\left( x\right) = {\sigma }^{r}\left( x\right) \neq x. \]\n\nFrom this we may conclude that \( {\left( \sigma \tau \right) }^{n} \) is not the identity. This completes the proof of (ii).\n\n\( {}^{4} \) These are the same exponentiation rules you saw in high school algebra: \( {x}^{ab} = {\left( {x}^{a}\right) }^{b} \) | 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 \( {a}_{j + 1} \) . | 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 \( {\left( f \circ g\right) }^{-1} = {g}^{-1} \circ {f}^{-1} \) for invertible functions \( f \) and \( g \) . Since transpositions are invertible functions, we have\n\n\[ {\left( \left( {a}_{1}{a}_{3}\right) \left( {a}_{1}{a}_{2}\right) \right) }^{-1} = {\left( {a}_{1}{a}_{2}\right) }^{-1}{\left( {a}_{1}{a}_{3}\right) }^{-1} = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \]\n\n(the second equality follows because every transposition is its own inverse.)\n\nIf we apply similar reasoning to the last three transpositions in the expression, we find\n\n\[ {\left( \left( {a}_{1}{a}_{4}\right) \left( {a}_{1}{a}_{3}\right) \left( {a}_{1}{a}_{2}\right) \right) }^{-1} = {\left\lbrack \left( {a}_{1}{a}_{3}\right) \left( {a}_{1}{a}_{2}\right) \right\rbrack }^{-1}{\left( {a}_{1}{a}_{4}\right) }^{-1} = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \left( {{a}_{1}{a}_{4}}\right) \]\n\nApplying this result inductively, we obtain finally:\n\n\[ {\mu }^{-1} = \left( {{a}_{1}{a}_{2}}\right) \left( {{a}_{1}{a}_{3}}\right) \cdots \left( {{a}_{1}{a}_{n - 1}}\right) \left( {{a}_{1}{a}_{n}}\right) , \]\n\nfrom this expression we may see that \( {a}_{1} \rightarrow {a}_{n},{a}_{n} \rightarrow {a}_{n - 1},{a}_{n - 1} \rightarrow {a}_{n - 2},\ldots ,{a}_{2} \rightarrow \) \( {a}_{1} \), which corresponds to the cycle we want. | Yes |
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