Notices
Results 1 to 33 of 33

Thread: Little question about groups

  1. #1 Little question about groups 
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Hi,

    Suppose you have a group of order 130, and suppose you were able to prove that it has exactly 13 different subgroups of order 2. Can you conclude that there are exactly 13 elements of order 2 (since they have the identity in common) or is it possible that another subgroup, say the subgroup of order 10 (which exists) contains an element of order 2 different than the 13?

    i.e. are they exactly 13 or more?

    Thanks very much


    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    In any group G, there is a 1-to-1 correspondence between elements of order two and subgroups of order two. The correspondence is given by:

    g |-> <g>

    Let's go through the details carefully.

    1. This map is 1-to-1: if I have two elements of order two g, h with <g> = <h>, then since <g> = {e,g}, <h> = {e,h}, and neither of g or h is equal to e, we must have g = h.

    2. This map is onto: if H is a subgroup of G of order two, let h be a nonidentity member of H. Then <h> is a subgroup of H of order two, hence <h> = H.

    So, in your case, you can conclude that there are 13 elements of order two.

    An interesting exercise would be to generalize this statement--what is the correspondence between elements of order n and subgroups of order n? The correspondence clearly cannot be 1-to-1 for n > 2: if g has order n, then so does g<sup>-1</sup>, and <g> = <g<sup>-1</sup>> while g ≠ g<sup>-1</sup>.


    Reply With Quote  
     

  4. #3  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Good idea, I didn't think about maps.

    Yes, it would be a very interesting generalization, but I don't think it's easy... in fact that what I was looking for.

    The idea is that, for no particular reason, I wanted to study the properties of non-abelian groups of order 130. And here is what I got:

    Let G a non-abelian group of order 130. Then
    1) G is not simple.
    2) G has normal subgroups of order 5, 13, 26, 65.
    3) The subgroups of order 5, 13 are unique.
    4) The subgroups of order 2, 5, 13 and 65 are cyclic.
    5) G is solvable, it has at least 3 decomposition series.
    6) The converse of Lagrange theorem holds for G.
    7) The subgroups of order 2 and 10 cannot be normal.
    8) There are 13 subgroups of order 2.
    9) The subgroup of order 10 has 13 conjugates.

    I think I now have everything I need about the subgroups, so I want to see if I can list the orders of the elements of G... or at least some of them.

    But I can't think of a way that relates the orders of subgroups to the orders of elements... mmm I'll think again, I'd better go study now; I have an exam tomorrow.

    Thanks a lot
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  5. #4  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I beleive I just found something significant, but I'm not sure...

    Consider the subgroup of order 65. All its elements must have order 1, 5, 13 or 65.

    Now, since the group is cyclic, it contains a subgroup of order 5 and 13. But these subgroups are unique. Hence, all the other elements must have order 65.

    That is, we now have 1 identity, at least 12 elements of order 13, at least 4 elements of order 5 and at least 48 elements of order 65.

    Now, consider the subgroup of order 26. All its elements must have order 1, 2, 13 or 26. From Sylow, this subgroup has a subgroup of order 13, which again is unique. Hence, there are 13 elements left, and these must have order 2 or 26. Now, since this subgroup is normal, then if it contains an element, it also contains all its conjugate. Hence, if it contains a 2, it contains all the 13 2. Now, if on the other hand it doesn't contain any 2, then all the 13 elements left must have order 26. Then this subgroup is cyclic. Then it must have elements of order 2 (contradiction).

    Hence, I think we have the following for now: Let's use the notation Hk for a subgroup of order k.

    H5 has 4 elements of order 5 + e
    H13 has 12 elements of order 13 + e
    H26 has 13 elements of order 2, 12 elements of order 13 + e
    H65 has 4 elements of order 5, 12 elements of order 13, 48 elements of order 65 + e
    Each of the 13 H2 of course has 1 element of order 2 + e.

    The trouble is for the 13 conjugates H10 for now. Each has a subgroup of order 5 which is unique. Hence, we have 4 elements of order 5 + e. There are 5 elements left. At least one of them must have order 2 from Sylow. This leaves 4 elements... would we get a contradiction if H10 was cyclic? Or maybe the 4 left should actually have order 10, then noticing that we have 13 conjugates for 10 and 13 elements of order 2 in total (so one in each), maybe we could find a link. I'm still thinking.

    But even if we get the structures of all subgroups, will we be able to do that classification? I think it's possible, but I really have to work now.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  6. #5  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I think I got it and it looks very beautiful, but I need your confirmation.

    G cannot have an element of order 130 or 26. 130 is clear, and 26 would make H26 cyclic, a contradiction from the above argument.

    So all elements of G must have order 1, 2, 5, 10, 13, 65.
    We know for sure that there is 1 element of order 1, 13 elements of order 2, 4 elements of order 5 and 12 elements of order 13. For 5 and 13, there cannot be more elements since these subgroups are unique.
    This leaves 52 elements. Since we don't have a choice, they must have order 10. Notice that this would make sense, for if you divide 52 by 13, you get 4, so that each H10 indeed seem to contain 4 elements of order 10 and 1 element of order 2. In fact, it looks like the 13 H2 are subgroups of the respective 13 H10.

    If what I'm saying is true, we have a very beautiful structure for any non-abelian group of order 130:
    A unique subgroup H5, H13. We have an H26 which seems to be unique since we only have 12 elements of order 13 and 13 elements of order 2 and 1 elements of order 1. Hence it seems there cannot be 2 different H26.
    Exactly 13 H2 and 13 H10 (are they really 13 H10 or more? we have 13 conjugates, but could we find another H10 other than these, not conjugate?)
    H2, H5, H10, H13, and H65 are cyclic. H26 is not cyclic, hence not abelian since it has the form pq with (p,q)=1.
    We have H1 normal in H5 normal in H65 normal in G
    H1 normal in H13 normal in H26 normal in G
    H1 normal in H13 normal in H65 normal in G
    Each of the 13 H2 is normal in each of the 13 H10, normal in G (note that each H2 is now normal coz H10 is cyclic)

    That makes 16 composition series. I think this takes care of it. Did I do something wrong?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  7. #6  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Actually, if what I said is true, can't we now say that there are only 2 groups of order 130 up to isomorphism: Z130 and D65?

    Coz, for non abelian groups, our isomorphism would just send each element of a given order to the one of the given order.

    What do you think?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  8. #7  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    I've come up with at least 4 nonisomorphic groups of order 130. The first two are the C<sub>130</sub> (cyclic groups) and D<sub>65</sub>, as you suggest. The next two are "semidirect products". For example:

    Consider the set C<sub>10</sub> x C<sub>13</sub>. Define a binary operation + on this set given by:

    (a,b) + (c,d) = (a+c,b+(-1)<sup>a</sup>d)

    where the +'s inside the parentheses are the usual group operation in cyclic groups. (-1)<sup>a</sup> means 1 if a is even and -1 if a is odd, which is a well-defined notion in any cyclic group of even order. This is a group:

    -(0,0) is the identity;
    -[(a,b)+(c,d)]+(e,f) = (a+c,b+(-1)<sup>a</sup>d)+(e,f) = (a+c+e,b+(-1)<sup>a</sup>d+(-1)<sup>a+c</sup>f) = (a,b)+(c+e,d+(-1)<sup>c</sup>f) = (a,b)+[(c,d)+(e,f)];
    -The inverse to (a,b) is (-a,-(-1)<sup>a</sup>b), as (a,b)+(-a,-(-1)<sup>a</sup>b) = (a-a,b-(-1)<sup>2a</sup>b) = (0,0).

    This group is nonabelian--for example:

    -(1,1)+(1,2) = (2,-1)
    -(1,2)+(1,1) = (2,1)

    But it has an element of order 10--namely, (1,0). D<sub>65</sub> only has elements of order 1, 2, 5, 13, and 65: thinking geometrically, rotations of a 65-gon must have order dividing 65, and reflections all have order 2. So this group is not on your list. We can similarly define a semidirect product on the set C<sub>26</sub> x C<sub>5</sub>, and this would give us a group with an element of order 26. And you can do some twiddling around to show these two groups are not isomorphic, I believe.

    Edit: fixed up some errors in bold above.
    Reply With Quote  
     

  9. #8  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    A little research suggests I've found all the groups of order 130. Here's a list of group orders (left column) against number of nonisomorphic groups (right column):

    http://www.research.att.com/~njas/sequences/b000001.txt

    The first kink in your reasoning was that the group of order 26 had to be normal. This is not the case in D<sub>65</sub>, and we can see this geometrically. D<sub>65</sub> contains a couple of groups of order 26. Draw a 65-gon and, starting with any vertex, draw a 13-gon by connecting each 5-th vertex starting with the one you chose. This gives us a 13-gon sitting inside of the 65-gon, and any symmetry of the 13-gon gives a symmetry of the 65-gon. Thus we have an embedding of D<sub>13</sub> into D<sub>65</sub>. There are 5 ways I can do this, so there are 5 such subgroups. They are all conjugate, as any element of order 2 is conjugate, D<sub>65</sub> has 65 elements of order 2, and D<sub>13</sub> only has 13.
    Reply With Quote  
     

  10. #9  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I see... but I don't know what's wrong with my proof...

    Here is how I proved that H26 must be normal:

    Let G act on the set of left cosets of H26 in G by left translation. Then the kernel K of the induced homomorphism T: G--> S5 must be a subgroup of H26.

    Now, from the 1st isomorphism theorem, G|K is isomorphic to T(G). And |T(G)| must be a divisor of |G| = 130 and |S5| =120. Hence |T(G)| = 2 or 5, hence |K| = 65 or 26. But |K| < or = 26 since it is a subgroup of H26. Hence the kernel K is all of the subgroup H26.

    But K is normal in G, hence H26 must be normal.

    What's wrong?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  11. #10  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    |T(G)| could be 10, in which case |K| could be 13. Evidently, this is the case when G = D<sub>65</sub>.
    Reply With Quote  
     

  12. #11  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Great, I can't even calculate...

    Thanks a lot; I'll be more careful next time before I send my new discoveries...
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  13. #12  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Ok, I had to give up my classification quest for some time because of my exams, but I've advanced much and I hope I'll be able to finish it before next exam (next Monday)...

    Here we are, it's almost done: (I'll denote by a(b) the fact that there are a elements of order b).

    For C130

    130 = 1(1) + 1(2) + 4(5) + 4(10) + 12(13) + 12(26) + 48(65) + 48(130)

    Beautiful symmetry, isn't it? (1,1 4,4 12,12 48,48... I wonder if it's just a coincidence)

    To prove this note that all subgroups are normal, hence the Sylow 2 is normal and unique implies 1(2), H10 must have 4(10) to complete itself, H26 must have 12(26) and H65 48(65) and the rest of the elements are of order 130 since it's cyclic.

    But as you can see there is an assumption I'm making here: that all our subgroups are unique. They are all normal, but are they unique? If they are not, then all is correct except the last step, namely, that there are 48(130). If these are not unique, there are less than 48(130).
    But I think we can fix this one: I remember last year there was a way to see all the generators of a cyclic group. I can't remember how though... Maybe that can help?

    For D65:

    130 = 1(1) + 65(2) + 4(5) + 12(13) + 48(65)

    I'm 100% sure about this one: here H26 is not normal, hence there cannot be 13(2). The choice is 5(2) or 65(2). However H65 has no elements of order 26, hence H26 has 13(2). Hence there must be 65(2). And the 48(65) are necessary to complete H65. No assumptions about this one, it's done.

    For C10 x C13

    130 = 1(1) + 13(2) + 4(5) + 52(10) + 12(13) + 48(65)

    This was actually the one I was describing as being the unique nonabelian group of order 130. The same reasoning as before. An assumption was made: again I'm assuming H65 is unique, so there must be 52(10) to complete the group. If H65 is not unique there are less than 52(10). Note that H26 is unique (as I said before), that there are exactly 13 subgroups of order 2, hence the single assumption is indeed that H65 is unique (in other words, that there are only 48(65)) Here it's a bit more troubling than C130 because our group is not cyclic. However, notice that H65 is cyclic and normal. Can that help?

    For C26 x C5

    The great trouble is about this one. I couldn't even make an assumption to list the elements. Here is what we have for sure:

    1(1) + 5(2) + 4(5) + 12(13) + 48(65)

    Where I'm assuming for the moment that there are 48(65) and not more.
    To prove 5(2) note that there cannot be 13(2) since H26 is not normal and there cannot be 65(2) since G has elements of order 26, so this would make too many elements.

    So, what about the elements of order 10, 26? I got the structures of the subgroups:

    H10 = 1(1) + 1(2) + 4(5) + 4(10)

    Again this strange symmetry. To prove this note that there cannot be 5(2) otherwise we would have a unique H10, which contradicts the fact that there are 13 conjugates H10. Hence there is at least 1(10). But then H10 is cyclic, hence there must be 1(2) and 4(10).

    H26 = 1(1) + 1(2) + 12(13) + 12(26)

    That symmetry cannot be a coincidence after all. It must be a property of some cyclic groups, don't you think?

    To prove this note that since G contains elements of order 26, then H26 is cyclic, then its Sylow 2 is unique. Also note that H26 has 5 conjugates

    So here is where I got. If there are indeed just 48(65) as I'm assuming, then there must be 60 elements of order 10 and 26. Note that we cannot have only 4(10) otherwise we couldn't account for the 13 conjugates of H10, since there are only 5(2). Hence there are at least 7(10) and 12(26).

    So without making any assumptions we have at least the following

    1(1) + 5(2) + 4(5) + 12(13) + 48(65) + 7(10) + 12(26)

    This leaves 41 elements unclassified which must split into elements of order 10 or 26 (again if there are only 48(65)).

    I don't know. Do you have an idea? Can you actually find the number of elements of order 10, 26 in that particular group you created? This might give us a hint, and since you say these are the only ones up to isomorphism, we can generalize the results.

    Thanks very much, I hope I'm not annoying you.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  14. #13  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    mmm maybe I made a mistake about the last one. I'm saying "since H10 is not normal", but for that statement I forgot I used my 1st wrong argument to prove it.

    Actually if H10 is normal the case is more beauitful. You have indeed all the 5(2) in H10 and 60(12) that split into 5 conjugate H26, just like the 52(10) splitted into 13(10).

    If that's true, then C26xC5 has no elements of order 10.

    If that's the case, then the clue is to prove that, any group of order 130 that have elements of order 26 has a normal subgroup of order 10. Then it would be done.

    I'll think about that.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  15. #14  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I'm trying to find a relation between the number of conjugates of a group G and the number of different conjugates of an element a in G.

    If we could just prove that if H10 has 13 conjugates, then the conjugates of a in H10 are either all one and the same element or 13 different conjugates then it would be done.

    That is, if G has n conjugates, what is the number of elements of the conjugacy class of a?

    If H10 is not normal, then it has 13 conjugates. There are 5 fixed elements in H10: 1(1) and 4(5). How many different conjugates can an element of order 2 or 10 have? If they both had 13 conjugates we would get too many elements (since we have elements of order 26). And if they both had each one and the same conjugate, then there would be a unique subgroup H10.

    I really feel that the following statement is correct:

    Let G a group of order 130. Then G has elements of order 26 <=> The subgroup of order 10 is normal.

    To prove <= note that if the subgroup of order 10 is normal, then it contains all the conjugates of an element a in G that have order 2. Hence there are either 1(2) or 5(2), then we must have elements of order 26 to complete H26.

    I'm trying to prove the more useful (for my case) =>
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  16. #15  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    The work for the first three groups looks good. I haven't looked at the fourth one--I'm kind of under the weather at the moment, and I'll get back to you once I'm up to it. Skimming your arguments, however, it looks like you've made some progress.
    Reply With Quote  
     

  17. #16  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I beleive I finally got it :-D I was able to prove my statement.

    Here we are:

    Suppose G has elements of order 26. Then H26 is cyclic, then its Sylow 2-subgroup is unique, then

    H26 = 1(1) + 1(2) + 12(13) + 12(13).

    There are 2 cases: H26 is normal in G or H26 is not normal in G (these are by the way the cases of C130 and C26xC5)

    If H26 is normal in G then it contains all the elements of G of order 2. Hence G has only 1(2). Hence G is cyclic (direct product of its Sylow subgroups) then H10 is normal. I had proved this one earlier, the problem was in the next case.

    If H26 is not normal, then it has 5 conjugates. Assume towards a contradiction that H10 is not normal. Then it has 13 conjugates.

    Now let G act on itself by conjugation, let a any element of order 26 and b any element of order 10.

    I'll denote the order of the conjugacy class of a by [a].

    Since [a] divides |G|, then [a] = 130, 65, 26, 13, 10, 5, 2 or 1
    Now note that any 2 a in H26 cannot be conjugates, for if that happens, then ga1g^-1 = ga2g^-1 ==> a1 = a2.
    Hence we have 12 different a that are not conjugate and we have 5 conjugates of H26.
    Hence [a] = 5, 2 or 1 which makes 60(26), 24(26) or 12(26).
    Since H10 is not normal, then it cannot contain all 5(2)

    (To prove 5(2) note that there cannot be 13(2) since H26 is not normal and there cannot be 65(2) since G has elements of order 26, so this would make too many elements.)

    Hence H10 contains at least 1(10). Then it is cyclic. Hence

    H10 = 1(1) + 1(2) + 4(5) + 4(10).

    Again these 4 are all different and not conjugates. Now

    [b] = 130, 65, 26, 13, 10, 5, 2 or 1

    130, 65 and 26 are too many while 1, 2, 5 are too few to account for 13 conjugates. Hence [b] = 13 or 10, which yields 52(10) or 40(10).

    Since we had only 60 remaining elements (we were sure about the first 70), then the 60 elements must be a combination of these number. The only possible choice is 60(26). Hence There are no elements of order 10, Hence H10 contains all elements of order 2 and is normal, a final contradiction.

    I'm glad. I am. Don't you say there is anything wrong!

    Hence the only part that needs to be confirmed (and I'm pretty sure it's correct in our case at least) is that the all the normal groups were unique, and that those that were not normal gathered all the elements of G.

    But wait a sec, doesn't that amount to say that conjugation makes a partition of G, which is correct?

    I beleive it's done. If it is I'll be really proud.

    Thanks a lot for your time. Even if what I said was correct, sharing it with you was very useful
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  18. #17  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    After writing 18 pages in "A paper on groups of order 130" I found a mistake in one of my proofs that is likely to make the whole paper fall apart.

    I cannot continue right now as I have to study. I'll try again next Monday, and if I don't finish it, my vacations are next Thursday anyway.

    Thanks again. You don't have to think about it (unless of course you'd like to); I want to do it myself. If you don't mind I may get your confirmation however.

    If you do mind (because you're likely to be busy) then I'll try with my scary Algebra professor. I like him but he's really scary.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  19. #18  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Okay, so my calculations show that there are no elements of order 10 in the fourth group. So there must be an error in your argument that there is at least one element (or subgroup) of order 10. In fact, I'm not quite sure how you concluded that H10 had to exist.
    Reply With Quote  
     

  20. #19  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    That's pretty simple: H5 is normal and there exists a subgroup of order 2 from Sylow theorem. Hence there is a subgroup H2H5. Since these intersect in <e> then H2H5 = H10. I don't think there is anything wrong with that.

    What you concluded about the 4th group (no element of order 10) is what I felt was true from symmetry, and that's why I proved that H26 is cyclic <=> H10 is normal.

    The bug I found is with G nonabelian + H26 is normal <=> there are 13 elements of order 2. And unfortunately the rest of the paper rests on that.

    But the results are so symmetric I really feel I'm correct but I need to fix my proofs. I'll try to do that today.

    The results again are the following:

    C130 = 1(1) + 1(2) + 4(5) + 4(10) + 12(13) + 12(26) + 48(65) + 48(130)

    D65 = 1(1) + 65(2) + 4(5) + 12(13) + 48(65)

    C10xC13 = 1(1) + 13(2) + 4(5) + 52(10) + 12(13) + 48(65)

    C26xC5 = 1(1) + 5(2) + 4(5) + 12(13) + 60(26) + 48(65)

    The reason why I say this is symmetric is that the 52(10) and 13(2) split into exactly 13 conjugate H10 and the 60(26) and 5(2) split exactly in 5 conjugate H26.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  21. #20  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    By the way, I know that if G is cylic and k divides |G| then there is a normal cylclic subgroup of order k.

    My question is: is it unique? Being normal sometimes implies being unique (like the Sylow subgroups). Is that true for the subgroups of a cyclic group?

    More generally, is it true if our subgroup is cyclic?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  22. #21  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I just proved that my more general question is false:

    In G = ZpXZp with addition, <(1,0)> is a normal cyclic subgroup of order p. However it is not unique: <(0,1)> is another normal cyclic subgroup of order p.


    However I think my special case question is true: If G is cyclic and k divides |G| then there is a unique normal cyclic subgroup of order k.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  23. #22  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    It looks like I'm not the only one who asked that question:

    http://www.physicsforums.com/archive.../t-167554.html

    The answer is yes, if G is cyclic then this subgroup is unique.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  24. #23  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Stranger
    That's pretty simple: H5 is normal and there exists a subgroup of order 2 from Sylow theorem. Hence there is a subgroup H2H5. Since these intersect in <e> then H2H5 = H10. I don't think there is anything wrong with that.
    Yeah, you're right. I was thinking for some reason that the group was cyclic, which it isn't.

    The results again are the following:

    C130 = 1(1) + 1(2) + 4(5) + 4(10) + 12(13) + 12(26) + 48(65) + 48(130)

    D65 = 1(1) + 65(2) + 4(5) + 12(13) + 48(65)

    C10xC13 = 1(1) + 13(2) + 4(5) + 52(10) + 12(13) + 48(65)

    C26xC5 = 1(1) + 5(2) + 4(5) + 12(13) + 60(26) + 48(65)
    These are correct. I calculated these directly from the explicit representations of the groups.
    Reply With Quote  
     

  25. #24  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Yeah, I've not been doing the best job of reading the details of your posts--for example, you had already concluded that there were no elements of order 10 in the fourth group when I made a comment about your prior statement that it did have such elements. In any case, it looks to me like you've probably worked everything out right--you definitely have the correct number of elements of each order, and I think you described the structure of the subgroups correctly, and this suggests that you must have calculated the number of conjugates of subgroups correctly.

    Some interesting things about groups:

    1. All subgroups of an abelian group are normal. Since all elements commute (and writing things additively), we have g+h-g = h for all elements, and so for any subgroup H, g+H-g = H. In particular, cyclic groups are abelian and hence all subgroups are normal. Also, all subgroups of cyclic groups are cyclic!

    2. Do you know about the structure of finite abelian groups? It turns out they're always a direct sum of finite cyclic groups. More generally, finitely generated abelian groups are direct sums of cyclic groups.

    3. We can partition the elements of a group by grouping them according to their order. However, we get an even finer and more important partition of the group if we look at "conjugacy classes"--i.e., let g be an element of g; then its conjugacy class is the set {hgh<sup>-1</sup> : h element of G}. Note that all elements of a conjugacy class have the same order, but the opposite need not be true. In particular, abelian groups can be characterized as groups whose conjugacy classes are all singletons.
    Reply With Quote  
     

  26. #25  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Thanks for your little notes

    And thanks for your confirmation. I can now proceed with more hope however, I left that paper for some time to write another paper about centralizers and normalizers, which will probably help me in my investigations.

    The problem is that I reached a contradiction at the end :? It is to be noted that I've never written anything correct till now
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  27. #26  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I don't wanna bother you but I reviewed the thing about centralizers many times and I can't find the error.

    Here is what I said: I'll denote by H^C(a) the intersection of centralizers of all elements in H, that is, the intersection of C(a) for all a in H. i.e., if g is in G, then g is in H^C(a) <=> ga = ag for all a in H. It's thus smaller than C(a) but larger than the center, and that's the point; I'm trying to generalize the properties of the center for this larger subgroup.

    Lemma 1
    Let G be a group and H a subgroup of G. If H is abelian then H is a subgroup of H^C(a)

    Proof: I beleive this one is trivially true. I put it here just in case... I began to doubt about everything I write.

    Lemma2
    Let G be a group and H a subgroup of G. Then H^C(a) is a normal subgroup of G.

    Proof:
    Let G act on H by conjugation. The kernel K of the induced homomorphism G-->A(H) is a normal subgroup of G. The induced homomorphism is given by g-->Tg where Tg:H-->H takes a--> gag^-1.
    Now, g is in K <=> Tg = I <=> Tg(a) = a for all a in H <=> gag^-1 = a for all a in H <=> g is in H^C(a)
    Hence H^C(a) is a normal subgroup of G.

    Lemma 3
    Let G be a group and H a p-subgroup of G. Then p divides |H^C(a)|.

    Proof:
    I'll use a lemma to prove this one. The lemma says:

    Let H be a p-group that acts on a finite set S. If S0 = {x in S| hx = x for all h in H} then |S| = |S0| (mod p) (I can't write the congruence)

    Now, let H act on G by conjugation. Then S = G and S0 = {g in G| aga^-1 = g for all a in H} = H^C(a). Hence from the lemma |H^C(a)| = |G| - kp. But that must divide |G|. Simplifying we get that |H^C(a)| = kp/ m-1 where m divides |G|. This cannot be unity. If you assume it is and note that p must divide |G| then you get p = 1/n-k where n divides |G|. Hence p divides |H^C(a)|

    Now lemma 2 and lemma 3 cannot be both correct. At least one of them is wrong. Consider a p-subgroup H of A5. Then from lemma 3 p divides |H^C(a)| and from lemma 2 H^C(a) is normal in A5. In order to avoid making A5 having a proper normal subgroup, we may deduce that H^C(a) = A5. However in that case H is a subgroup of the center of A5 and H is a proper normal subgroup of A5.

    There is another problem In C26xC5 I say that H26 is not normal. However, since H26 is cyclic, the using lemma 1 H26 is contained in H26^C(a). But the order of the latter divides |G|. Hence |H26^C(a)| is either 26 or 130. If it's 26 then H26^C(a) = H26 is a normal subrgoup using lemma 2. And if it's 130 then H26 is a subgroup of the center, then H26 is normal in G.

    So what do you think?
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  28. #27  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Stranger
    Lemma 1
    Let G be a group and H a subgroup of G. If H is abelian then H is a subgroup of H^C(a)

    Proof: I beleive this one is trivially true. I put it here just in case... I began to doubt about everything I write.
    This is right--by definition, the elements of H^C(a) are those that commute with elements of H. Since H is abelian, all of its elements commute with each other.

    Lemma2
    Let G be a group and H a subgroup of G. Then H^C(a) is a normal subgroup of G.

    Proof:
    Let G act on H by conjugation...
    This action only makes sense if H is normal! The lemma is false as stated, too. Let G = A<sub>n</sub>, n ≥ 5, and let H = <(12)(34)>. Then H^C(a) is nontrivial--it contains (12)(34), but it doesn't contain, for example, (12)(45), as (12)(45) (12)(34) (12)(45) = (12)(35). Since G is simple, we have H^C(a) is not normal.

    Lemma 3
    Let G be a group and H a p-subgroup of G. Then p divides |H^C(a)|.

    Proof:
    I'll use a lemma to prove this one. The lemma says:

    Let H be a p-group that acts on a finite set S. If S0 = {x in S| hx = x for all h in H} then |S| = |S0| (mod p) (I can't write the congruence)

    Now, let H act on G by conjugation. Then S = G and S0 = {g in G| aga^-1 = g for all a in H} = H^C(a). Hence from the lemma |H^C(a)| = |G| - kp. But that must divide |G|. Simplifying we get that |H^C(a)| = kp/ m-1 where m divides |G|. This cannot be unity. If you assume it is and note that p must divide |G| then you get p = 1/n-k where n divides |G|. Hence p divides |H^C(a)|
    The idea here is completely right, although I got lost in your arguments (specifically, I don't see how you concluded that if kp/(m-1) = 1, then p = 1/(n-k); nor do I see how, if |H^C(a)| > 1, then p must divide |H^C(a)|). The argument I see is a little simpler: we know |H^C(a)| = |G| (mod p) from the sublemma, and we know that |G| = 0 (mod p), as p divides the order of H and hence of G. Thus, by transitivity of congruence modulo p, we have |H^C(a)| = 0 (mod p), which literally means p divides |H^C(a)|.

    Now lemma 2 and lemma 3 cannot be both correct. At least one of them is wrong. Consider a p-subgroup H of A5. Then from lemma 3 p divides |H^C(a)| and from lemma 2 H^C(a) is normal in A5. In order to avoid making A5 having a proper normal subgroup, we may deduce that H^C(a) = A5. However in that case H is a subgroup of the center of A5 and H is a proper normal subgroup of A5.

    There is another problem In C26xC5 I say that H26 is not normal. However, since H26 is cyclic, the using lemma 1 H26 is contained in H26^C(a). But the order of the latter divides |G|. Hence |H26^C(a)| is either 26 or 130. If it's 26 then H26^C(a) = H26 is a normal subrgoup using lemma 2. And if it's 130 then H26 is a subgroup of the center, then H26 is normal in G.
    Yeah, both of these problems stem from the incorrect statement of lemma 2.
    Reply With Quote  
     

  29. #28  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    Well I didn't know it had to be normal in fact, we only studied action of groups on sets. But then in some proof we made the group act on itself, so I assumed it was ok. Thanks a lot for this information.

    But still, this means that the result is not entirely wrong: if the subgroup H is normal, then H^C(a) is also normal, which is not a bad result after all.

    As for lemma 3, I didn't say then p must divide but since p must divide. Here is what I meant: assume on the contrary that this is unity. Then m = pk + 1. But m = |G| / (|G| - pk). Hence |G| = pk + 1. But p divides |G|. Hence pk + 1 = np for some n. Then p = 1/(n-k). But surely your argument is simpler.

    I even found another way to prove this by making use of a more familiar result: since H is a p-subgroup then p divides the order of its center. But its center is contained in H^C(a), hence p divides |H^C(a)|.

    As for the other paper, I finally fixed everything using very simple ideas. Even the proof I wrote here, in which I proved my "statement" had some bugs. But everything is ok now, I got the orders of all elements (which I showed you), and I finally concluded by giving the class equation of each group. I just have to write it down.

    So now I basically have eveything I need, but I was thinking: can I find the generators? That's the only thing I didn't get. Of course I know the generators of D65 (and subgroups isomorphic to it), but, do you have an idea how can I discover them? given the fact that I finally found the center and conjugacy classes of each group (I'll show you everything when I'm done with writing it, but you don't have to read it, for it's long. You may skip everything and go to the results if you want).

    Thanks again.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  30. #29  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Stranger
    Well I didn't know it had to be normal in fact, we only studied action of groups on sets. But then in some proof we made the group act on itself, so I assumed it was ok. Thanks a lot for this information.
    You understand why a G can act on a subgroup H by conjugation iff H is normal in G, yes?

    I even found another way to prove this by making use of a more familiar result: since H is a p-subgroup then p divides the order of its center. But its center is contained in H^C(a), hence p divides |H^C(a)|.
    This is probably the most elegant way of doing things! It appeals to the "center of a p-group is nontrivial" result, which is well known.

    ...but I was thinking: can I find the generators? That's the only thing I didn't get.
    The semidirect product groups can be generated by two elements. For example, C<sub>26</sub>xC<sub>5</sub> can be generated by (1,0) and (0,1). You can see this by noting that (1,0) generates all elements of the form (a,0), and if you have an element of the form (a,b), then (a,b)-(a,0) = (0,b). Then note that (0,1) generates elements of the form (0,b), and you're done. The same idea works for the other semidirect product.

    ----------------------

    Let me say a little bit about semidirect products...

    We can form general semidirect products as follows: let N and H be groups, and suppose there is a homomorphism from H to the automorphism group Aut(N). In other words, h acts on N via automorphisms. Then we define the semidirect product:

    G = HxN = {(h,n): h in H, n in N}

    And the group law is:

    (h,n)(g,m) = (hg,n h(m))

    where h(m) is h acting on m. You can go ahead and check that this satisfies the group laws.

    Now note that G contains subgroups isomorphic to H and N--namely, the elements (h,e) and the elements (e,n). Identify H and N with these subgroups. Then their intersection is e, and they generate the whole group: we have (e,n)(h,e) = (h,n) (as e must act trivially on N). And N is a normal subgroup--the explicit calculations require calculating inverses, but note that the inverse of an element (h,n) must be of the form (h<sup>-1</sup>,*), where * is some undetermined element, and a product (h,n)(g,m) = (hg,*), where again * is some undetermined element. Then conjugation gives:

    (g,m)<sup>-1</sup>(e,n)(g,m) = (g<sup>-1</sup>,*)(g,*) = (e,*)

    And this is an element of N. So, summarizing, if G = HxN is a semidirect product, then H and N are subgroups of G with trivial intersection, H and N generate G, and N is normal. In general, if you can find two subgroups of a group G, one of which is normal, whose intersection is trivial, and which generate G, then G is the semidirect product of these two groups.
    Reply With Quote  
     

  31. #30  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    You must have guessed I didn't study semidirect products I didn't study the "nilpotent" thing or the "splitting" thing either, things I come across as I read. But I'm planning to study them during this vacation since we're starting Ring Theory next semester. I have around 2 weeks, seems more than reasonable.

    Ok, I get most of what you mean I beleive, I just have to read more about it. So if I understand well, we could say that D65 is the semidirect product of H65 and H2, C26xC5 the semidirect product of H5 and H26 while C10xC13 is the semidirect product H13 and H10. Indeed, each pair contains a normal subgroup and intersect in <e>.

    mmm, maybe the tricky part is the "generate G" condition. How did you know that H65 and H2 would not generate C10xC13 for instance? I mean, how can you predict the internal structure of a semidirect product, given the internal structure of the two groups?

    Another question: you say

    if you can find two subgroups of a group G, one of which is normal, whose intersection is trivial, and which generate G, then G is the semidirect product of these two groups.
    And if both were normal, wouldn't we get a direct product? I'm just making sure of my informations.

    Well, I don't want to annoy you with my "thanks" all the time, so I'll save you a very big thanks when I finish that paper
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  32. #31  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I forgot about the action in my first post. Well, I think G can act on a subgroup H by conjugation iff H is normal in G because this action can be defined as a map from GxH--->H that takes (g,h) to ghg^-1. Hence, if H is not normal, ghg^-1 may not be in H and the action is indeed meaningless.

    PS I made 2 posts in case you didn't notice. I'm saying that because the 2nd post starts a new page, so that you may not notice the first.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

  33. #32  
    Forum Professor serpicojr's Avatar
    Join Date
    Jul 2007
    Location
    JRZ
    Posts
    1,069
    Quote Originally Posted by Stranger
    So if I understand well, we could say that D65 is the semidirect product of H65 and H2...
    Good eye! In general, you can always realize D<sub>n</sub> as a semidirect product of a normal cyclic subgroup of order n and a cyclic subgroup of order 2.

    mmm, maybe the tricky part is the "generate G" condition. How did you know that H65 and H2 would not generate C10xC13 for instance? I mean, how can you predict the internal structure of a semidirect product, given the internal structure of the two groups?
    That's a good question. I honestly don't know how you can tell whether two semidirect products will give the same group. I guess you just have to check...

    And if both [subgroups of a semidirect product] were normal, wouldn't we get a direct product? I'm just making sure of my informations.
    Also correct! Say H and N are normal subgroups of G with trivial intersection and G = NH. Then you can show nh = hn for any n in N, h in H, and so you can find an isomorphism f: NxH -> G by f(n,h) = nh.

    I forgot about the action in my first post. Well, I think G can act on a subgroup H by conjugation iff H is normal in G because this action can be defined as a map from GxH--->H that takes (g,h) to ghg^-1. Hence, if H is not normal, ghg^-1 may not be in H and the action is indeed meaningless.
    You're batting one hundred! That's exactly it. Group actions are very useful and pop up all over the place; you just gotta make sure your actions are well-defined.
    Reply With Quote  
     

  34. #33  
    Forum Sophomore Stranger's Avatar
    Join Date
    Sep 2005
    Posts
    148
    I'm finishing the paper. I'm in the last few pages. I generalized it and fixed everything. If you could send me a mail on

    tadnid@gmail.com

    for me to know your address (or write it down here if you don't mind) I would send it to you.


    Thanks very much


    I'm gonna mention you in my acknowledgment :-D

    The only reason why I'm not writing the paper here is that I used mathtype in it, which doesn't want to appear here, and I'm too lazy to switch all the symbols by funny substitutes.
    Watch what thy eyes can't see... and live it.

    (T.B)
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •