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How many homomorphisms are there

  1. May 15, 2006 #1

    JFo

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    How many homomorphisms are there from [itex]S_5[/itex] to [itex]\mathbb{Z}_5[/itex]?

    Well there is at least one, the trivial homomorphism, ie: every element of S_5 gets mapped to 0.

    I have a feeling that this is the only homomorphism but am having trouble proving that no other homomorphism could exist. Any suggestions?

    I know that every nonzero element of Z_5 has order of 5, and for a non-trivial homomorphism to exist, there needs to be some element of S_5 (not equal to the identity) that gets mapped to a nonzero element of Z_5.
     
    Last edited: May 15, 2006
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  3. May 15, 2006 #2
    I believe that there are other homomorphisms. Think about the subgroups of S_5. Does it have any cyclic subgroups of order 5?

    SBRH
     
  4. May 15, 2006 #3

    Hurkyl

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    But that only gives you a map in the other direction: from the cyclic group into the permutation group.

    What are the generators of S_5?
     
  5. May 15, 2006 #4
    Yeah, yeah, yeah...

    I just realized my silly mistake and came back to announce its sillyness!

    Ignore my previous reply.

    SBRH
     
  6. May 15, 2006 #5

    JFo

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    Im thinking the generators of S_5 are all the transpositions of S_5. Is this correct?

    Im working out a proof, here's what I have so far:

    proof:
    Let [itex] \phi : \mbox{S}_5 \rightarrow \mathbb{Z}_5[/itex] be a nontrivial homomorphism. Now [itex] |\phi(\mbox{S}_5)| [/itex] divides [itex]|\mathbb{Z}_5|[/itex], and thus must be 1 or 5. It can't be 1 since phi is nontrivial and thus must be 5, hence phi is onto.
    Let H = Ker([itex]\phi[/itex]). We can form the factor group [itex]\mbox{S}_5 / H[/itex] which is isomorphic to [itex]\mathbb{Z}_5[/itex]. This implies that [itex]|H| = 24 [/itex]. So for each element x of Z_5 there are 24 distinct elements of S_5 that map to x.

    Ok Im stuck again
     
    Last edited: May 15, 2006
  7. May 15, 2006 #6

    Hurkyl

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    Yes. Every element of S_5 is a product of transpositions.


    Are there any 24-element subgroups of S_5? Normal ones?
     
    Last edited: May 15, 2006
  8. May 15, 2006 #7

    JFo

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    No. There are no 24-element subgroups of S_5 that are normal. Thanks much for your help Hurkyl!
    I was wondering.. is there a way, say if your looking at a subgroup diagram of S_n for some n, to tell whether a given subgroup of S_n is normal?

    Also, is there a way to figure out which subgroups are conjugate without computing every conjugation of that group?
     
  9. May 15, 2006 #8

    Hurkyl

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    Incidentally, looking at the generators is just as easy. What is the image of any transposition?

    (I would have said easier, but I suppose if you really understand this stuff, knowing that there are no 24-element subgroups of S_n is probably of comparable difficulty)


    Isn't it the case that A_n is the only normal subgroup of S_n, for n > 4?
     
  10. May 16, 2006 #9

    JFo

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    I suppose if [itex] \tau [/itex] is a transpostion, then [itex] \tau^2 = i[/itex], thus [itex]\phi(i) = \phi(\tau^2) = \phi(\tau)^2 = 0[/itex]. This implies [itex]\phi(\tau) = 0[/itex] since there are no nonzero elements of order 2 in Z_5. Ahhh... this then implies that that all of S_5 gets mapped to 0 since every permutation can be written as a product of transpositions and phi is a homomorphism. Is this the correct reasoning?

    You're right, this is an easier solution.
    Thanks again for your help!
     
    Last edited: May 16, 2006
  11. May 16, 2006 #10

    Hurkyl

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    Yes, that is the line of thinking I originally had. (Until I was inspired by your computation of |H|)
     
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