anonymous
  • anonymous
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 2-cycles
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
How do you feel about part a?
anonymous
  • anonymous
S_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
  • anonymous
proving that involutions in S_n are precisely products of disjoint 2-cycles is easy enough in the product of disjoint 2-cycles => involution direction, since they tell you that all 2-cycles are involutions. it is slightly less easy to show the other direction, namely that an arbitrary involution is a product of disjoint 2-cycles

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Perhaps 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
  • anonymous
I do not fully know how to complete part b
anonymous
  • anonymous
Do 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 2-cycles \(\sigma_1\cdots\sigma_n\). 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 2-cycles 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
  • anonymous
Okay, to find the converse do I assume a disjoint cycle and show that it is an involution?
anonymous
  • anonymous
yes, exactly
anonymous
  • anonymous
I 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
  • anonymous
that's only one way
anonymous
  • anonymous
Okay. SO that is saying that we have an involution sigma^2=(1) and leading to disjoint 2 cycles?
anonymous
  • anonymous
maybe it would help if you did it the other way too. S o I can see the entire picture
anonymous
  • anonymous
that is saying that a product of disjoint 2-cycles is an involution. to go the other way, suppose we have some permutation \(\sigma\). by cycle decomposition there is a more-or-less (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
  • anonymous
That makes more sense now. Thanks a ton!

Looking for something else?

Not the answer you are looking for? Search for more explanations.