How to find a formula for the probability that two elements chosen at random generate the symmetric group Sn?
Q. are and pi = 3.14 same thing
Asked by YAHOO_USER__ - Sun Jun 15 10:23:44 2008 - - 1 Answers - 0 Comments

A. Try where it says Netto's conjecture states that the probability that two elements and of a symmetric group generate the entire group tends to 3/4 as . This was proven by Dixon (1969). The probability that two elements generate for , 2, ... are 1, 3/4, 1/2, 3/8, 19/40, 53/120, 103/168, ... (Sloane's A040173 and A040174). Finding a general formula for terms in the sequence is a famous unsolved problem in group theory.
Answered by a_math_guy - Sun Jun 15 13:50:23 2008

How many conjugacy classes are there in Symmetric group 4.?
Q. I'm going to guess 4, because the conjugacy class is defined as the orbit of the set of elements that conjugate to each other... obviously a permutation which permutes the same number of elements can be commuted to by some other permutation of the same number of elements ... so the number of different conjugacy classes should just be the number of different permutation sizes? Correct?
Asked by mt_rand - Sun May 4 03:53:56 2008 - - 2 Answers - 0 Comments

A. it's the number of partitions of 4, which map to the length of cycles in the conjugacy class. for each partition, i'll list a member of the conjugacy class. 4 (1234) 31 (123) 22 (12)(34) 211 (12) 111 () = e
Answered by holdm - Sun May 4 04:16:56 2008

Prove that the transpositions (1 2), (2 3),..., (n-1, n) generate the symmetric group Sn?
Q. Prove that the transpositions (1 2), (2 3),..., (n-1, n) generate the symmetric group Sn
Asked by jin_nzzang - Sun Oct 25 10:50:20 2009 - - 1 Answers - 0 Comments

A. HINT: (12)(23)(12) = (13) (13)(34)(13) = (14) ... Steve
Answered by stephenmdalton - Tue Oct 27 08:20:30 2009

Why is the symmetric group on a finite set, S, guaranteed to not be cyclic if it has more than 2 elements?
Q. Why is the symmetric group on a finite set, S, guaranteed to not be cyclic if it has more than 2 elements?
Asked by Nick - Tue Sep 16 21:31:24 2008 - - 1 Answers - 0 Comments

A. It is not abelian. Steve
Answered by stephenmdalton - Fri Sep 19 10:25:33 2008

absract algebra: show that the symmetric group S4 is solvable?
Q. solvable if there exists a chain G0,G1,G2...Gr of groups such that G0=G,Gi+1 is a normal subgroup with prime index in Gi for 0<=i<=r-1 note orderGi/order Gi+1 must be a prime. this is my problem which groups (index) must be prime?
Asked by special k - Sat Oct 18 17:47:56 2008 - - 1 Answers - 2 Comments

A. Cyclic groups of prime order are the only groups of prime order. The point is, to compute Gi / G(i+1), you just look at |Gi| / |G(i+1)|. This works for any setting in which the original group is of finite order - obviously, if Gi and G(i+1) are infinite, there's not really anything like that. But that's another story... So if these quotients are prime, then you're fine. So work it out: s a < (12)(34) , (13)(24) > c c c 1 Computing: |S | / |A | = 24/12 = 2 (prime) |A | / |< (12)(34) , (13)(24) >| = 12/4 = 3 (prime) |C C | / |C | = 4/2 = 2 (prime) |C | / |1| = 2/1 = 2 (prime) That's the composition series proving S is solvable.
Answered by cheeser1 - Sat Oct 18 18:41:38 2008

Did you know that the symmetric difference of sets makes the power set a group?
Q. Given the set X, consider the set of its subsets, P(x). Consider the following operation in P(X): A B = (A-B) U (B-A). Prove that P(X) with the operation is a Group. Have fun. Prove using this fact that the only prime factor of the cardinality of P(X) (assuming X is finite) is 2.
Asked by ghijk3 - Mon Jun 30 08:27:14 2008 - - 1 Answers - 0 Comments

A. i) Just go down the list of conditions that need to be satisfied: 1) Associativity, (A B) C = A (B C): This is the only "tricky" step, but can easily be done by using Venn diagrams. It is a mess to write down (especially in Yahoo Answers), so I will leave it to you to finish. Everything else is pretty easy: 2) Identity: Oe, the empty set, is the identity element. A Oe = (A\Oe) U (Oe\A) = A U Oe = A. 3) Inverse elements: Every element is its own inverse. A A = (A\A) U (A\A) = Oe U Oe = Oe. 4) Closed under : This is clear, since (A\B) U (B\A) is a subset of X, if A and B are. ii) This is easy using Cauchy's Theorem: "If p, a prime, divides the order of a group G, then there exists an element g in G such that the order of g is p." … [cont.]
Answered by Eulercrosser - Mon Jun 30 09:34:47 2008

Find the number automorphism of the symmetric group of order 3?
Q. Find the number automorphism of the symmetric group of order 3?
Asked by sarkar_malay_bir - Tue Apr 25 11:00:37 2006 - - 1 Answers - 0 Comments

A. S3 is isomorphic to the dihedral group of order 6, which consists of maps over the equilateral triangle with unique sides. There are only two basis elements, the flip over a central axis, and a single rotation. All of the other 4 maps can be written as products of these two actions. Since a homomorphism is completely defined by its action on the basis, we have 2! automorphisms (ways of mapping the basis into itself).
Answered by Ron - Tue Apr 25 11:27:51 2006

how to find all subgroups of the symmetric group S3 ?
Q. how to find all subgroups of the symmetric group S3 ?
Asked by qadir - Mon May 28 10:08:03 2007 - - 2 Answers - 0 Comments

A. I've been working on similar problems recently. I didn't find a solution explicitly for S3, but I did find one for D6 which is isomorphic to S3, so we're still okay. I'm going to list all subgroups. 1) S3 2) {Identity} 3) {Identity, (abc), (acb)} = A3 4) {Identity, (ab)} 5) {Identity, (ac)} 6) {Identity, (bc)} So to my best knowledge, there are 6 subgroups of S3. Note: Some sources refer to D6 as D3. It's an unfortunate non-standardized notation. I prefer D6, so that's what I used.
Answered by unknown - Mon May 28 11:34:20 2007

prove that the symmetric group S3 is indecomposable.?
Q. prove that the symmetric group S3 is indecomposable.?
Asked by quynheagle - Mon Mar 3 11:04:22 2008 - - 1 Answers - 0 Comments

A. If you mean show that there are not nontrivial subgroups H and K such that G=H x K, then you have to first see if there are normal subgroups H and K whose orders multiply to the order of S3, which is 6. We would need normal subgroups of orders 3 and 2. There is a normal subgroup of order 3 that could be H, but there is NO normal subgroup of order 2, so there is no K that could go with the H.
Answered by okdan - Mon Mar 3 20:53:22 2008

Show that the symmetric group , S_n , is non-Abelian for n > or = 3 .?
Q. Show that the symmetric group , S_n , is non-Abelian for n > or = 3 .?
Asked by Randolph G - Sun Oct 4 10:36:09 2009 - - 2 Answers - 1 Comments

A. Your group Sn will contain (12) and (23). Compute: (12)(23) = (132) Compute: (23)(12) = (123) which are not the same. Thus Sn is not Abelian.
Answered by cheeser1 - Tue Oct 6 11:09:55 2009

symmetric group permutation?
Q. if a= ( 1 2 3 4 5 6 7 8 9 ) ( 4 3 5 7 2 8 9 6 1 ). How can i find |a^-1000|
Asked by lena - Mon Nov 9 13:09:18 2009 - - 1 Answers - 0 Comments

A. Writing this permutation in cycle notation yields a = (1 4 7 9) (2 3 5) (6 8). (If you're not familiar with this notation, it means that 1 4 7 9 1, 2 3 5 2, and 6 8 6. Hopefully this makes sense...) Since these cycles are independent, it is the case that a^(-1000) = (1 4 7 9)^(-1000) (2 3 5)^(-1000) (6 8)^(-1000). Since the first cycle has order 4 and the second has order 2, and 4 and 2 are both factors of -1000, each of these will be the identity. Thus, a^(-1000) = (2 3 5)^(-1000). Since 3 is NOT a factor of -1000, the 3-cycle (2 3 5) will NOT be eliminated; therefore, |a^(-1000)| = 3.
Answered by jeredwm - Mon Nov 9 15:33:32 2009

abstract algebra: what does the notation using the little square mean? is it a group?
Q. show the symmetric group S3 is solvable. (chain of sub groups) with prime index G0=G,G(i+1).
Asked by special k - Sun Oct 19 13:06:06 2008 - - 1 Answers - 0 Comments

A. I don't have a single clue what the little square you're talking about is. However, here is the easiest solvable series for S3 I can think of: S3 > A3 > {e}
Answered by Awms A - Sun Oct 19 13:40:53 2008

Help with hard group theory (Sylow subgroups)?
Q. Let p be an odd prime. First, find a set of generators for a p-Sylow subgroup K of S_p^2 (the symmetric group with degree p^2). Then find the order of K and determine whether it is normal in S_p^2 and if it is abelian. What have gotten so far is that the order of the subgroup is p^(p+1). But how do I do the other parts? Thanks.
Asked by A A - Wed May 13 18:31:57 2009 - - 1 Answers - 0 Comments

A. The generators of a Sylow group (the others Sylow groups are conugate to it) are: (1,2,3,...,p), (p+1,p+2,...,2p), ... (p^2-p+1,...,p^2), and (1,p+1,2p+1,...,p^2-p+1) (2,p+2,2p+2,...,p^2-p+2) ... (p,2p,3p,...,p^2). it's not normal (in fact An are the only normal subgroups of Sn when n>4) The generators commute, so the group is abelian I will add more details next days
Answered by Mielu istetz - Thu May 14 21:44:30 2009

Define what it means for a group to be cyclic?
Q. Q,1 Define what it means for a group to be cyclic? For a group to be cyclic, is a group that can be generated by a single element a. Can i answer this question with this definition? Q2, is Z11|{0} under multiplication modulo 11 is cyclic. I know it's cyclic because 11 is prime. Z|{0} p is prime it's cyclic. but how do we prove it? one more, is the symmetric group s4 is cyclic?
Asked by Charles - Sat Sep 5 21:42:44 2009 - - 2 Answers - 0 Comments

A. You might want to rephrase, as your grammar is a little awkward, but otherwise, your answer to Q1 works. For Q2, show it is cyclic by finding an element that generates the group. For instance, try 4: 4^1 = 4 4^2 = 5 4^3 = 9 4^4 = 3 4^5 = 1 so 4 doesn't generate the group. Try this with other elements until you find an element that does generate Z11\{0}. As for S4: it is not cyclic. It is easy to show that every cyclic group is abelian. Show that there are two elements in S4 that don't commute and you've shown S4 is not cyclic. Otherwise, you could go through all 24 elements and argue why none generates S4 itself.
Answered by Awms A - Sat Sep 5 21:57:01 2009

1)prove:the order of a permutation is the least common multiple of the orders of the cycles which make it up?
Q. question 2: prove that the transpositions (1 2),(2 3),...,(n-1,n) generate the symmetric group S_n.
Asked by greenheaven - Fri Oct 19 19:04:09 2007 - - 1 Answers - 0 Comments

A. (1) Hint. Let p be a permutation that acts on {1,...n} and let d(i) be the size of the cycle of p containing i, i=1,2,...n. Let M be the LCM(d(1),d(2),...,d(n)). (a) Prove that if p^M' = identity, then M' is divisible by d(i) for all i=1,...,n, and hence, must be divisible by M. (b) Prove that p^M = identity. That shows that M is the order of p. (2) (a) Prove it is true for n=1,2. (b) Use induction: Given any p in S_(n+1), we first want to write p as in the form: p = t1.t2.t3...tk.p' where ti's are transpositions of this sorrt, and p' has the property that p'(n+1)=n+1. Then use the induction step to show that p' can be written as transpositions of this sort.
Answered by thomasoa - Sat Oct 20 18:41:32 2007

need help with mathematical proof?
Q. show that the functions f=1/x, g=(x-1)/x generate a group of functions, the law of composition being composition of functions, which is isomorphic to the symmetric group S_3.
Asked by greenheaven - Sat Oct 13 18:03:52 2007 - - 2 Answers - 0 Comments

A. I agree with what Curt said. And the missing element is g^2, because it is the inverse of g..:) then mapping with S_3. Find the ones that has the same order, and map them.
Answered by Kitty M - Mon Oct 15 02:45:04 2007

Number Theory and Algebra Theory?
Q. 1. Prove by induction, for n=1,2,3...that 1^3 +2^3+3^3+...+n^3=((n(n+1) )/2)^2 2. Is the symmetric group S(subscript 3) isomorphic to Z(subscript 6), the group of inteers modulo 6 with addition (mod 6) as its binary operation? Give reasons for your answer 3. State what is meant by a subgroupe of a group 4. Is the dihedral group D(subscript 8) isomorphic to Z(subscript 8), the group of integers with addition modulo 8 as its binary operation? Give reasons for your answer. 5. Prove that if G and H are cyclic groups of order 2007, then G and H are isomorphic I need full working out and explinations because this is study for my exam and these are questions that im having real trouble on. I really appreciate it. Thanks Don
Asked by thegr8don - Fri Oct 12 03:09:26 2007 - - 3 Answers - 0 Comments

A. 1. 1^3 + 2^3 + ... + n^3 = n^2(n+1)^2/4 The prove is going by induction First for n =1 we have 1 = 1^2(1+1)^2/4 which is true Now suppose it's true for some n, then 1^3 + 2^3 + ... +n^3 + (n+1)^3 = n^2(n+1)^2/4 + (n+1)^3 = (n+1)^2(n^2 + 4n + 4)/4 = (n+1)^2(n+2)^2/4 QED 2. S_3 is NOT an Abelian group and Z_6 IS an Abelian group. That means they are not isomorphic 3 (H,*) is the subgroup og group (G,* if and only if (i) H is a subset of G (ii) if a and b are from H then a*b is in H (iii) if a is in H then a^-1 is in H 4. The same as 2 D_8 is not abelian Z_8 is abelian They are not isomorphic 5. If G is a cyclic group of order 2007 then G = {e = a^0, a, a^2, ...a^2006} for some a in G The same way H = {f = b^0, b, b^2, ...b^2006}… [cont.]
Answered by Ivan D - Fri Oct 12 04:26:11 2007

Isomorphism?
Q. 3.Prove that any group G is isomorphic to a certian subgroup of Sg where Sg is a symmetric group generated by G.
Asked by a5b14 - Fri Apr 11 01:44:53 2008 - - 1 Answers - 0 Comments

A. Any element g of G generates a permutation of the elements of G by sending any element, h, say, of G to gh. Call this permutation p_g. It's then easy to verify that {p_g | g in G} is a subgroup of the group of permutations of G and is isomorphic to G via the map g -> p_g.
Answered by nudnik0 - Fri Apr 11 04:00:36 2008

Permutations?
Q. In the symmetric group S9 consider the permutation p= (13742)(173)(5867). Determine the order of p. Write p as a product of transpositions. Find a permutation q in S9 with q^4 = p or show that no such q can exist. Would I be right in saying the order is 12 , the transpositions are (1,4)(4,2)(5,8)(8,6)(6,7) . The last part however not a clue
Asked by Terry C - Thu May 29 20:40:52 2008 - - 1 Answers - 0 Comments

A. p = (1 4 2) (5 8 6 7) The order is indeed 12. The transpositions also look okay, though I might say that (5 7)(5 6)(5 8) and (1 2)(1 4) might be preferable ways to write the cycles. For your question, note that p is composed of an odd number, 5, of transpositions. If q existed, how many transpositions (odd/even) would q^4 be made of? Good luck.
Answered by K-Dub - Fri May 30 13:00:33 2008

How to destroy America....Please read and state your opinions?
Q. "How to Destroy America!" This One is very thought provoking! We know Dick Lamm as the former Governor of Colorado. In that context his thoughts are particularly poignant. Last week there was an immigration overpopulation conference in Washington, DC, filled to capacity by many of American's finest minds and leaders. A brilliant college professor by the name of Victor Hansen Davis talked about his latest book, Mexifornia, explaining how immigration - both legal and illegal - was destroying the entire state of California. He said it would march across the country until it destroyed all vestiges of The American Dream. Moments later, former Colorado Governor Richard D Lamm stood up and gave a stunning speech on how to destroy… [cont.]
Asked by wranglergal_2 - Sun Dec 24 13:01:00 2006 - - 20 Answers - 1 Comments

A. The Coming Darkness A Warning for Americans: A message from South Africa 20th March 2004 People used to say that South Africa was 20 years behind the rest of the Western world. Television, for example, came late to South Africa (but so did pornography and the gay rights movement). Today, however, South Africa may be the grim model of the future Western world, for events in America reveal trends chillingly similar to those that destroyed our country. (Read what happened in S. Africa at the following website~it's where we are headed if we don't heed the warnings. The story is down the page. "The Coming Darkness")
Answered by CherokeeRose - Sun Dec 24 14:30:10 2006