How do i find the conjugacy classes of Dihedral group of order 8?
Q. I know Dih8 = {1, a, a^2, a^3, b, ab, (a^2)b, (a^3)b} I have been told that the conjugacy classes are: {1} , {a^2}, {a,a^3} , {b,(a^2)b} and {ab,(a^3)b} But i have no idea how to get these...
Asked by Paul W - Tue Oct 20 15:37:56 2009 - - 1 Answers - 0 Comments

A. First, the center of Dih8 is {1,a^2}, so these elements have their own conjugacy class: {1}, {a^2}. Now bab = a^3, so a^3 is conjugate to a. Since these are the only elements of order 4 (and conjugation preserves order) they make up another class: {a, a^3}. You can check the others, conjugating each by a and b. Steve
Answered by StephenD - Thu Oct 22 08:53:54 2009

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

Are the sizes of the conjugacy classes of Sn the same?
Q. How do you determine this? I would have liked it if they were... would make designing my algorithm a bit easier... but of course they probably aren't.
Asked by mt_rand - Fri May 23 03:49:06 2008 - - 1 Answers - 0 Comments

A. Nope. For instance, the conjugacy class of cycles of length n has (n-1)! elements. Meanwhile, the conjugacy class of cycles of length (n-1) (and one fixed point) has n*(n-2)! elements
Answered by Mo - Fri May 23 03:58:01 2008

Explain the structure of conjugacy classes of D_n, distinguishing carefully?
Q. between the cases where n is even and where n is odd. ___ Read off the centre of D_n from the calculation of above. You should find that Z(D_n) is {e} when n is odd and {e,r^n/2} when n is even. any help is appreciated.
Asked by moviegoer - Mon Nov 16 09:30:00 2009 - - 1 Answers - 0 Comments

A. This link should help you with this problem: I hope this helps!
Answered by kb - Mon Nov 16 15:04:18 2009

Probability of two elements of a group commuting?
Q. I'm being really dumb tonight I think...my brain isn't working. Any help on this problem: Let G be finite and have exactly k conjugacy classes. Show that the probability that two randomly chosen elements of G commute with each other is exactly k/|G|. The second element is to be randomly chosen after the fist is replaced.
Asked by MathNerd - Mon Oct 13 22:08:46 2008 - - 1 Answers - 0 Comments

A. The probability is equal to p = |C(g)| / |G|^2, where C(g) is the centralizer of g (i.e. the set of elements that commute with g). Let Cl(g) denote the conjugacy class of g. Then |Cl(g)| = [G:C(g)] = |G| / |C(g)|. Let H_1, H_2, ..., H_k denote the conjugacy classes of H. Then p = sum(g in G) |C(g)| / |G|^2 = sum(i=1 to k) sum(g in H_i) |C(g)| / |G|^2 = sum(i=1 to k) sum(g in H_i) 1/(|Cl(g)| * |G|) = sum(i=1 to k) sum(g in H_i) 1/(|H_i| * |G|) = sum(i=1 to k) 1/|G| = k/|G|.
Answered by Low Key Lyesmith - Mon Oct 13 23:33:27 2008

Let G be a group & let [x] denote the conjugacy class of x in G. Show | [g] | = |[g inverse]| for any g G.?
Q. Let G be a group & let [x] denote the conjugacy class of x in G. Show | [g] | = |[g inverse]| for any g G.?
Asked by han - Mon Mar 31 11:37:59 2008 - - 1 Answers - 0 Comments

A. This is FALSE. For a counter example, take any abelian group with an element g of order 3 or greater (such as a cyclic group of order 3 or more). In abelian groups every conjugate class contains exactly one element, but g=g^(-1) if and only if g^2=e (identity element).
Answered by okdan - Tue Apr 1 07:26:19 2008

Find all normal subgroups of D2n (dihedral) by treating the cases where n is odd and even separately.?
Q. Consider the conjugacy classes in D2n
Asked by chamomilla - Wed Oct 14 14:37:30 2009 - - 1 Answers - 0 Comments

A. Terminology: The dihedral group is generated by a rotation R and a reflection F: R^n = e F^2 = e Elements of the form R^k are "rotations." Elements of the form R^k F are "reflections." --- --- --- --- --- --- --- --- --- --- Known properties of the dihedral group: Any element of the dihedral group can be written R^k or R^k F for 0 k[cont.]
Answered by cheeser1 - Sat Oct 17 03:56:41 2009

Please help with a group theory proof?
Q. Say G is a finite group. Prove that if some conjugacy class of G has exactly two elements then G cannot be simple. Thanks!
Asked by A A - Wed May 13 23:27:16 2009 - - 1 Answers - 0 Comments

A. Given a finite group G, and an element g, the size of g's conjugacy class is the same as [G:C(g)], the index of its centralizer. Thus if g's conjugacy class has size 2, C(g) has index 2 in G, so is normal. Of course, G is then not simple as long as |G| > 2. But if |G| 2 it is abelian, so every conjugacy class has 1 element. Steve
Answered by StephenD - Fri May 15 02:28:46 2009

Questions about group algebras?
Q. I have two questions and I need help badly! Let G be any group (written multiplicatively) and let F be any field. Construct the F-algebra with G as a basis and with multiplication defined by extending the original group multiplication on G on the whole algebra. This F-algebra, denoted FG, is called the group algebra of G over F. 1. Show that the group algebra FG contains a finite-dimensional ideal iff G is finite. Show that if G is finite, then there is a one-dimensional ideal. 2. Let F be a field and G a finite group. Show that the center Z(FG) is an F-subspace of the group algebra FG with dimensional equal to the number of conjugacy classes in G. Thanks for your help!
Asked by olgratefuldead - Mon Feb 5 00:01:39 2007 - - 1 Answers - 0 Comments

A. 1. If G is finite you take J the ideal who has basis one element( in other words, the ideal generated by) g = g_1 + g_2 + ... + g_n, where g_i are all elements of G. The ideal is one-dimensional, for example elemets in the ideal are of the form fg, where f is in FG, so f = sum f_i g_i, where f_i are in F and g_i as above, sum f_i g_i h = sum f_i g_ i ( sum g_j) = But g_i sum g_j = sum g_j, this is important, it means that the product with one element permutes the elements of the group, or in other words g'G=G, for any g'. This property is satisfied only by finite groups. So in the sum above you get = sum (sum f_i) g_i = sum f_i {1} * sum g_i, which means that is J generated over F by g, i.e. J is one-dimensional ideal. As for… [cont.]
Answered by Mielu istetz - Mon Feb 5 01:31:19 2007

Hard group theory help?
Q. What are the conjugacy classes in S5 and how do I find them? Secondly, if two elements are conjugate in S5, how do I know when they are conjugate in A5? Thanks.
Asked by Kalinka - Thu Mar 19 12:49:09 2009 - - 1 Answers - 0 Comments

A. For your first question, conjugacy in S5 preserves cycle structure and every two elements with the same cycle structure are conjugate. Thus, you have 7 conjugacy classes: (a b c d e) (a b c d) (e) (a b c) (d e) (a b c) (d) (e) (a b) (c d) (e) (a b) (c) (d) (e) (a) (b) (c) (d) (e) - identity element
Answered by abelian - Thu Mar 19 13:12:17 2009

Prove the L_2 (8) group is simple.?
Q. The group l_2 (8) has 504 elements, and has nine conjugacy classes of order 1, 56, 56, 63,72,72,72. Prove the L_2 (8) group is simple.
Asked by sagar - Wed Dec 2 20:30:52 2009 - - 1 Answers - 0 Comments

A. You left off two 56's. A normal subgroup is the union of conjugacy classes. You just need to show no sum from the set {1, 56, 56, 56, 56, 63, 72, 72, 72} (and including 1) is a divisor of 504 = 2^3 * 3^2 * 7. You need to check 2^8-2 = 254 possibilities. A computer might help. In Mathematica, Table[ 1+Apply[Plus, i], {i, Subsets[ {56,56,56,56,63,72,72,72} ] } ] will list all possibilities. By the way, L_2(n) where n>3 is always simple. Steve EDIT - actually, there is no need to check so many possibilities since alot of your numbers are the same. It seems there are only 40 possibilities. I got this in Mathematica with A:=Table[ 1+Apply[Plus, i], {i, Subsets[ {56,56,56,56,63,72,72,72} ] } ]; B:= DeleteDuplicates[A]; B
Answered by StephenD - Fri Dec 4 03:31:21 2009

Please help me with this question testings your ability to describe symmetries geometrically etc?
Q. This question tests your ability to describe symmetries geometrically and to represent them as permutations in cycle form. It also tests your understanding of conjugacy classes and their relationship to normal subgroups. The figure for this question is prism with three congruent rectangular faces and an equilateral triangle at each end. The locations of the vertices of the prism (numbered 1 2 3 on one triangle and 4 5 6 on the other) have been numbered so that we may represent the group G of all symmetries of the prism as permutations of the set {1,2,3,4,5,6}. a) Describe geometrically the symmetries of the prism represented in cycle form by (14)(26)(35) and (13)(46). b) Write down all the symmetries of the prism in cycle form as… [cont.]
Asked by Timi - Sat Jul 18 06:48:04 2009 - - 1 Answers - 0 Comments

A. (14)(26)(35) is 2-fold rotation about an axis perpendicular to the 3-fold axis that runs through the centers of the triangular faces. The 2-fold rotation axis bisects the 1-4 edge of the prism and runs through the center of the opposite rectangular face. (13)(46) is a reflection operation through a mirror plane that contain the 2-5 edge of the prism and contains the the 3-fold axis of symmetry. b-c) There are 12 operations that form 6 conjugacy classes: (I) The identity, (1)(2)(3)(4)(5)(6); (II) two 3-fold rotations - clockwise and counterclockwise, (123)(456) and (132)(465); (III) three 2-fold rotations - the one described in (a) plus (25)(16)(34) and (36)(15)(24); (IV) a single reflection operation, (14)(25)(36); (V) two "improper [cont.]
Answered by Timothy - Tue Jul 21 01:04:14 2009

This question concerns the symmetry group G of the regular hexagon?
Q. Regular hexagon corners numbered 1 to 6 anti-clockwise. Let g in G be the anticlockwise rotation of the hexagon through an angle of 4pi/3 about its centre, and let h in G be the reflection of the hexagon in the line through the midpoints of sides 1 6 and 3 4. a) Write g, g^2 and h in cycle form b) Express the conjugate ghg^-1 of h by g in cycle form, and describe ghg^-1 geometrically c) Are (14)(23)(56) and (14)(25)(36) conjugate in G? Justify the answer d) Are (14)(23)(56) and (14)(25)(36) conjugate in S_6? Justify the answer e) Write down the conjugacy class of G that contains (14)(23)(56) Please help if you can Best answer given
Asked by Timi - Wed Sep 16 16:39:01 2009 - - 1 Answers - 0 Comments

A. a) A rotation of 2pi/6 = pi/3 moves each vertex one place over: (1 2 3 4 5 6). So, a rotation moves each vertex 4 spaces: g = (1 5 3)(2 6 4). g^2 = (1 3 5)(2 4 6). h = (16)(34)(25). b) g h g^(-1) = (1 5 3)(2 6 4) * (16)(34)(25) * (1 3 5)(2 4 6) = (14)(23)(56). First, we rotate clockwise 4pi/3, then we reflect (over the line specified above), and finally rotate counterclockwise 4pi/3. c) No; the first is a reflection, while the second is a rotation. d) Yes, they are of the same cycle type. e) Compute x * [(14)(23)(56)] * x^(-1) for all x in G. Let f = (14)(23)(56). I believe it comes out to {f, r^2 f, r^4 f} where r is a rotation of 2pi/6. (You can double check that...). I hope that helps!
Answered by kb - Wed Sep 16 17:19:09 2009