Answers to problems2

: 136.330 Intro to Algebra: Answers to problems2
   By Paul on Tuesday, October 29, 2002 - 07:46 pm: Edit Post

This is what the proof says:

To prove that the order of An is 1/2(n!) it suffices to prove that Sn has the same number of even permutation as odd permutations, since Sn has order n!. To do this, it suffices to prove that the mapping theta: An to Sn defined by theta(a) = a(1 2) is one-to-one and that theta(An) is the set of all odd permutations in Sn.

I don't understand the entire last sentence.

Thank you


   By Sasho on Tuesday, October 29, 2002 - 08:47 pm: Edit Post

I am going to use f instead of theta. They define a mapping f:An -> Sn as follow: take any even permutation a and define f(a) to be the composition (the product) a(12), where (12) is the two cycle sending 1 to 2 and 2 to 1. Since a is even (product of even many 2-cycles), the product a(12) is an odd permutation (since it is a product of odd many 2-cycles - the ones you use for a - and the 2-cycle (12).) This (essentially) shows that f(An) is the set of all odd permutations (i.e, we have covered the last part of the last sentence in the statement that you give above). If we show the f is 1-1 that will show that An has as many elements as f(An); that is, we would have that the number of even permutations is the same as the number of odd permutations. Since each permutation is either odd or even, it follows that the set of all permutations is twice larger then the set An of even permutations. Since the former has n! elements, it follows that the latter has 1/2(n!) elements, as stated.

The whole story will be completed if we show that f is indeed 1-1. I am leaving that part for you. Feel free to ask for hints if you have problems with it.


   By Paul on Monday, November 11, 2002 - 10:12 pm: Edit Post

Hello, I was just wondering if you could help me solve questions 14.19 and 15.20 from the text. Here they are...

14.19 Assume m, n are in Z. Find necessary and sufficient conditions for (the subgroup generated by m) subset (the subgroup generated by n).

15.20 Prove that if A is a subgroup of G and B is a subgroup of H, then A x B is a subgroup of G x H.

Thanks


   By Sasho on Monday, November 11, 2002 - 10:46 pm: Edit Post

14.9. The subgroup of Z generated by m is made of all multiples of m. So, the subroup generated by m is a subset of the subgroup generated by n if m is a multiple of n.

15.20. I am going to denote the operations by the "invisible" multiplication sign, or by ".".

(a) Multiplication closure. Take two elements of AxB, say (a1,b1) and (a2,b2), and multiply them: (a1,b1).(a2,b2)=(a1a2,b1b2) (this is the way the operation works in AxB - componentwise). Since A is a subgroup (of G), A is closed under multiplication in G, and so a1a2 is in A (since both a1 and a2 are in A). Similarly, b1b2 is in B. So (a1a2,b1b2) is in AxB.

(b) Closed under taking inverses: similar story as in (a).

p.s. please use "create new coversation" button more often.


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