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Group/Representation Theory Help! (Summations and elements of a finite group)

  1. Jun 21, 2011 #1
    Apologies, the LaTeX thing doesn't seem to be working, so not very clear!

    I am working through a book on representation theory, but am stuck on these exercises.

    1. The problem statement, all variables and given/known data
    Let G be a finite group and let E = Sum g (running over all g in G).

    (i) Prove: Ex = E (forall x in G)
    (ii) Prove: E^{2} = nE (|G| = n)
    (iii) Find 3 e_{i} in G s.t. (e_{i})^{2} = e_{i} (forall i = 1,2,3)

    2. Relevant equations
    Unknown


    3. The attempt at a solution
    (i) In order to show that Ex = E, we must show that there is a bijection that maps from G to G which sends g to gx for all g in G. Must show that this map is surjective and injective to show it is bijective. Little bit stuck on showing these.

    (ii) I am told that part (ii) requires part (i). Am I correct in thinking that E^{2} = (Sum {running over elements of G} (Sum {running over elements of G} g))? Basically the sum of the sum! (Sorry LaTeX working would make this far easier to explain!)

    (iii) Again, part (iii) requires part (ii), but I'm not even quite sure what I'm being asked to show. Is it simply that there are 3 elements in the finite group G that when squared are equal to themselves?

    Thanks for any help.
     
  2. jcsd
  3. Jun 21, 2011 #2

    micromass

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    Hi OMM! :smile:

    First, can you tell me what book you're reading? Question 3 makes little sense to me, so I must be misunderstanding something. I'd like to read it from the original source.

    Indeed, you must show that

    [tex]\phi:G\rightarrow G:g\rightarrow xg[/tex]

    is bijective. For injectivity, you must show that

    [tex]\phi(g)=\phi(g^\prime)~\Rightarrow~g=g^\prime[/tex]

    or thus

    [tex]xg=xg^\prime~\Rightarrow~g=g^\prime[/tex]

    Can you show this??

    Do something analogous for surjectivity.

    So you'll need to calculate

    [tex]\left(\sum_{g\in G}{g}\right)\left(\sum_{g^\prime\in G}{g^\prime}\right)=\left(\sum_{g\in G}{g\left(\sum_{g^\prime\in G}{g^\prime}\right)}\right)[/tex]

    Now apply (1).

    If you don't see something immediately, then it might be worth to consider a little example like [itex]\mathbb{Z}_2[/itex]...
     
  4. Jun 21, 2011 #3
    Hi it's "Representations & Characters Of Groups" by Liebeck & James. Although the questions aren't from there, they were set for me to help me understand irreducible CG submodules and CG algebras a bit better! As I'm struggling here on this!!! Aghhh!



    Thanks, I can see the injectivity solution. Simply multiply on the left of both sides by x^{-1}, which is an element of G as x clearly is. And thus you get the required one-to-one property.

    For surjectivity, do I need to show that there's an element, which when it has [tex]\phi[/tex] applied to it, I get g? i.e. 1/x?


    Clearly here we're setting [tex]\left(\sum_{g^\prime\in G}{g^\prime}\right) = x[/tex] and it is running over the n elements of G, so we're applying Ex = E "n times" to get E^{2} = n.E?


    Thanks for your help, I'll consider using this simplification, to see if it makes things clearer!
     
  5. Jun 21, 2011 #4

    micromass

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    Fix g. What value y must I take such that xy=g?


    Yes, that's basically it. Here is the same argument in symbols:

    [tex]E^2=\left(\sum_{g\in G}{g}\right)E=\left(\sum_{g\in G}{gE}\right)=\left(\sum_{g\in G}{E}\right)=|G|\cdot E[/tex]
     
  6. Jun 22, 2011 #5
    I managed to get a bit more information about part (iii) and I was told to use a specific example of S_3, the symmetric group of order 6.

    (iii) asks to find 3 solutions to the equation (e_i)^{2} = e_i inside the group algebra C[S_3].

    where C = complex numbers.

    I guess the identity element is clearly 1 such element. However, finding the other 2 appears to be a mystery to me!!!
     
  7. Jun 22, 2011 #6

    micromass

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    Well, you've found out that [itex]e_1=1[/itex] satisfies the criterium.
    Now, what if I take [itex]e_2=E[/itex], then we have [itex]e_2^2=E^2=nE=ne_2^2[/itex]. This doesn't satisfy what we want, but perhaps we can modify our [itex]e_2[/itex]so that it does give what we want.

    For example, we could take [itex]e_2=cE[/itex] with c a certain complex number. What complex number could we take such that [itex]e_2^2=e_2[/itex]?
     
  8. Jun 22, 2011 #7
    Thanks for your help! After a bit of trial and error, I can see we could take c = 1/n and that would satisfy for e_2!

    So we now have [itex]e_1 = 1[/itex] and [itex]e_2 = (1/n)E[/itex]

    For [itex]e_3[/itex] I assumed we'd take a complex conjugate of "c" or (1/-n), but that obviously won't work as we end up with: [itex]e_3^2=-e_3[/itex]. Or am I barking up the wrong tree?
     
  9. Jun 22, 2011 #8

    micromass

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    What about taking the zero element?
     
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