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Group Theory problems

  1. Mar 7, 2006 #1

    AKG

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    1.
    (a) Find the number of p-Sylow subgroups in G = GL(n,Fp).
    (b) For two p-Sylow subgroups H and J of G, consider the number of elements in the intersection. What number appears this way for given p and n?
    (c) For a given p-Sylow subgroup H of G, find the number of p-Sylow subgroups with the given number of elements in the intersection with H.

    2.
    (a) Describe all groups of order 9.
    (b) Describe all groups of order 27.

    3. Find the table of irreducible characters for A4.

    4. Find the dimensions of irreducible representations of S4.

    5. For the non-trivial 1-dimensional irreducible representations of A4, find the character of the induced representation of S4, and determine if this representation is irreducible.

    I have some idea how to do these questions (or at least some of them). I'll post some work a little later when I get time. But although I have some idea as to what to do, I don't really have any idea as to what's going on. I think right now I'm at the point that I'm trying to get comfortable with the definitions and basic results by applying them to actual problems, so I hope if someone can help me with the above problems, I can get a clearer picture of the underlying concepts.
     
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  3. Mar 8, 2006 #2

    matt grime

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    4. is the easiest. Work out the number of conj classes of S_4, that's the number of integers you're looking for. S_4 obviously has triv, sign and the natural 3-d perm rep from quotienting out trivial submodule from S_4's action permuting a basis of C^4. Now the rest is straight forward (if you know about tensors, 3d tensor sign is another character), since the sums of squares of the dimensions are the size of the group (24), and all dimensions divide the order of the group.

    5 is just calculation from 3 and 4 using inner products
     
  4. Mar 8, 2006 #3

    matt grime

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    If you want a good reference then Reperesentations and Characters of Groups by James and Liebeck is good, really good. Contains all character tables of all groups you might need (ie all things of order less than 500 in one way or another)

    Remember, all groups of order p^2 are abelian (p a prime) and all p-groups are soluble.

    Actually, finding the complete character table of S_4 is very easy assuming you know about the inner product on characters.
     
  5. Mar 9, 2006 #4

    AKG

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    Okay, to see if I'm getting this stuff, I gave myself a similar problem:

    Find the full character table of S3. Find the induced representation in S4, and determine if it's irreducible

    We look at the possible cycle structures for elements of S3, and we see there are three conjugacy classes corresponding to e, (12), and (123). The trivial and sign characters give two 1 dimensional characters. Since the sum of the dimensions of the representations have to give the order of the group, we know that the final character corresponds to a 2-dimensional representation. So it sends the conjugacy class of e (i.e. {e} itself) to 2. We know that S3 has a natural action on the triangle, so we can look at a representation that maps S3 to the matrices in GL(2,R) that do the appropriate things to the triangle, and compute their traces. In fact, we only have to do it for (12) and (123). We get 0 and -1 respectively. We compute the inner product of this character with itself to see we get a value of 1, thereby ensuring that the character is irreducible? Well, that's what my TA told me, but something about that sounds fishy.

    The irreducible characters are supposed to form an ONB of the space of class functions. Let X and Y be two such characters. Then 2-1/2(X + Y) should be a normal vector, since:

    [tex]\langle 2^{-1/2}(X + Y),\, 2^{-1/2}(X + Y)\rangle[/tex]

    [tex]= \frac{1}{2}\langle (X + Y),\, (X + Y)\rangle[/tex]

    [tex]= \frac{1}{2}\left (\langle X,\, X\rangle + \langle X,\, Y\rangle + \langle Y,\, X\rangle + \langle Y,\, Y\rangle\right )[/tex]

    [tex]= \frac{1}{2}(1 + 0 + 0 + 1) = 1[/tex]

    since {X, Y} should form an orthonormal set. So we get a new vector of norm 1, which, according to my TA means that it is an irreducible character, so if we call it Z, then {X, Y, Z} should form an orthonormal set, but it clearly doesn't because Z is a linear combination of X and Y. Moreover, the Z we get from this is different from the character we get by looking at the symmetries of the triangle. Anyways, looking at the one we get from the symmetries of the triangle, it is easy to see that not only is it normal, but it is also orthogonal to the other two characters, so maybe this is enough to say that it's irreducible.

    Moving on, we look at the 2-dimensional character and see what we get when we induce a character in S4. Again, we see the conjugacy classes are:

    e, (12), (123), (12)(34), and (1234)

    The character [itex]\chi[/itex] induced from [itex]\chi _0[/itex] is defined by

    [tex]\chi (g) = \frac{1}{|S_3|}\sum _{x \in S_4} \chi ^o (xgx^{-1})[/tex]

    where [itex]\chi ^o(xgx^{-1}) = \chi _0 (g)[/itex] if xgx-1 is in S3, and 0 otherwise.

    We immediately see that [itex]\chi[/itex] will map (12), (12)(34), and (1234) to 0. It's also immediate that it will map e to 8. Now there are a number of ways to see that this will not give an irreducible character. First, if we already know the character table of S4, we know it has no 8-dim characters, so this isn't one. Also, we know that 8² = 64 > |S4|, but the order of the group must be the sum of the squares of the dimensions, so this can't be irreducible. Finally when we calculate the norm of this character, we see we'll get something greater than or equal to 64/24 which is greater than 1, so this character is not normal, and hence not an irreducible character.

    Anyways, we just have to compute what the character does to (123). First, we see this conjugacy class has 8 elements, 2 of which are in S3 (namely (123) and (321)). We want to know, for how many x in S4 do we get x(123)x-1 in S3. Well it's easy to show that if H is the group of stabilizers of (123), and t is any three cycle, then if y is such that y(123)y-1 = t, then every element z such that z(123)z-1 = t is of the form yh, for some h in H. Now the order of the group is 24, and the order of the orbit of (123) under conjugation is 8, so its stabilizer has order 3. From the above argument, there are 3 things which send (123) to (123) under conjugation, and 3 things which send it to (321). In fact, for any three cycle t, there are 3 things which send it to t. However, we're only concerned with the number that keep (123) in S3. We've counted 6, so the formula for the induced character, plus the fact that the original character sent (123) to -1 implies that the induced character sends it to -1 as well (i.e. it's -1 x 6 [the number of elements x such that x(123)x-1 is in S3] x 1/6 [since 6 = |S3|]). So to summarize, we had an irreducible character in S3 which sent e, (12), and (123) to 2, 0, and -1 respectively. This induced a character which sent e, (12), (123), (12)(34), (1234) to 8, 0, -1, 0, and 0 respectively.

    There are three "problems" with the induced character:

    - it is not normal (i.e. <X,X> is not 1)
    - it is of dimension 8, but if it were normal, then the sum of the squares of the dimensions would be greater than the order of S4
    - we know the table for S4, and this character isn't one of them [in particular, no character in the table sends e to 8]

    Now the main problem I have is:

    When can we tell if a character is irreducible?

    It's a necessary condition that it be normal. My TA says its sufficient. I think I might have made a mistake earlier when I said it sounded fishy. I constructed another class function which was normal. But arguably it wasn't a character (i.e. it wasn't the trace of some representation of the group, it was just some function that happened to be constant on the conjugacy classes of the group). So is it right? That if X is a character, then it is sufficient for it's norm to be 1 to say that it is irreducible?
     
  6. Mar 10, 2006 #5

    matt grime

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    it's acceptable, but not as elegant as it should be. S_3 has a natural action on any 3-d space by permuting a basis, e_1,e_2,e_3, and obviously e_1+e_2+e_3 is a subrep, and the quotient is 2-d, its character of an element g is what you wrote.

    If G is any group acting on any finite set X, then there is an obvious action of G on a vector space (over any field) with basis indexed by x in X, the character on g in G is no. points fixed by g.



    no, no, no. it is a class function of norm 1, it is not a character of a rep of G at all.


    And as for your last question, then, yes, if X is a character, then X is irreducible if and only if it's norm is one.

    The simplest way to see your construction is not a character is to observe that its value at the identity, which is the 'dimension' of the vector space on which you're attempting to assert this corresponds to a G-action is not rational, never mind an integer.


    It's just elementary Fourier analysis: every rep (over C) is a direct sum of simples:


    [tex] V= \sum_{\chi \in {\rm Irr}(G)} n_{\chi}S_{\chi}[/tex]

    where S_{\chi} is the irreducible rep with character chi, and the n_{\chi} are integers, and you can work out its norm, and its irreducible iff exactly one of the n_{chi} is 1 and the others are zero. n_{chi} is the inner product of X_V with chi. a class function corresponds to a character if and only if its inner product with any simple character is a positive integer (and I think 0 is a positive integer, though it is not a strictly positive integer).


    I saw nothing in what you wrote that backed up the assertion that you think your TA siad that a class function of norm one was in irred char. all i saw was him say that if X was a char, and its norm was one then ti was irred, which is certainly true.


    Anyway, the induced character you worked out is the sum of the two 3-d irreps of S_4 plus the 2-d irrep of S_4, if that's of any interest.
     
    Last edited: Mar 10, 2006
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