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anonymous
 one year ago
Let S_n denote the group of all permutations of the set [n]={1,2,...,n}. A permutation sigma in S_n is called an involution if sigma^2 =1, where 1=(1) in S_n denotes the identity. Explicitly list all evolutions of S_1, S_2, S_3, S_4. Fore general n, show that a permutation 1 not equal to sigma in S_n is an involution if and only if sigma is a product of disjoint 2cycles
anonymous
 one year ago
Let S_n denote the group of all permutations of the set [n]={1,2,...,n}. A permutation sigma in S_n is called an involution if sigma^2 =1, where 1=(1) in S_n denotes the identity. Explicitly list all evolutions of S_1, S_2, S_3, S_4. Fore general n, show that a permutation 1 not equal to sigma in S_n is an involution if and only if sigma is a product of disjoint 2cycles

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do you feel about part a?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0S_n are sufficiently simple for \(n\in\{1,\dots,4\}\) to manually list the involutions. an involution is essentially just a permutation that is its own inverse, i.e. it undoes itself when applied a second time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0proving that involutions in S_n are precisely products of disjoint 2cycles is easy enough in the product of disjoint 2cycles => involution direction, since they tell you that all 2cycles are involutions. it is slightly less easy to show the other direction, namely that an arbitrary involution is a product of disjoint 2cycles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Perhaps you can begin by considering for each of the sets of permutations S_1, S_2, S_3, S_4. \[ [1] = \{1\} \\ [2] = \{1,2\} \\ [3] = \{1,2,3\} \\ [4] = \{1,2,3,4\}.\] Then \[S_1 = \{1\} \\ S_2 = \{1,(1,2)\} \\ \vdots\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do not fully know how to complete part b

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you recognize that disjoint cycles commute with one another? Suppose then we have a permutation \(\sigma\) that can be factored as a product of disjoint 2cycles \(\sigma_1\cdots\sigma_n\). It follows then that $$\sigma^2=\sigma_1\cdots\sigma_n\sigma_1\cdots\sigma_n=\sigma_1^2\dots\sigma_n^2=e^n=e$$since 2cycles are obviously involutions, i.e. \(\sigma_1^2=\sigma_2^2=\dots=\sigma_n^2=e\), the identity; since they are disjoint, we were able to commute the \(\sigma_1,\dots,\sigma_n\) so that they square to the identity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, to find the converse do I assume a disjoint cycle and show that it is an involution?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am sorry but I am still confused with what you have written. Would you mind explaining it to me. Is that just one way or both ways of the proof.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay. SO that is saying that we have an involution sigma^2=(1) and leading to disjoint 2 cycles?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe it would help if you did it the other way too. S o I can see the entire picture

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that is saying that a product of disjoint 2cycles is an involution. to go the other way, suppose we have some permutation \(\sigma\). by cycle decomposition there is a moreorless (i.e. up to commutativity) unique representation in products \(\sigma=\sigma_1\cdots\sigma_n\). if it's an involution we have \(\sigma^2=e\implies\sigma=\sigma^{1}\) so $$\sigma_1\cdots\sigma_n=\sigma_n^{1}\cdots\sigma_1^{1}$$note the cycles are disjoint we can commute freely. do you see a way to deduce that \(\sigma_i=\sigma_i^{1}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes more sense now. Thanks a ton!
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