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Sylow p-subgroups: Group of order 693

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data


    I am having problems with parts c and d


    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2

    micromass

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    For (c). You have an action that sends

    [tex]\phi:C\rightarrow Aut(A)[/tex]

    with [itex]\phi(c):A\rightarrow A:a\rightarrow cac^{-1}[/itex].

    What is the kernel of this action?? Use that [itex]C/\ker(\phi)[/itex] is a subgroup of Aut(A).

    For (d). You know that A is a subset of C(a) (show this). You know that B is a subset of C(a). And you know that C(a) contains an element of order 7 (what does this imply for the order of C(a)?)
     
  4. Dec 2, 2011 #3
    ok, so for (c) I know that the kernel has order either 1 or 11, since it is a subgroup of C. If it has order 11, we are done because then every automorphism is the identity. I'd like to show that have order 1 leads to contradiction. If it has order 1, then |C/kernel|=11. But since the number of permutations of A is 9!, and 11 does not divide 9!, this leads to contradiction. Thus, the order of the kernel is 11 and every automorphism is the identity and fixes every element of A. Is this right?

    For (d), if A is a subgroup of C(a) and B is a subgroup of C(a), AB is a subgroup of C(a) and |AB|=99 since |A intersect B|=1. Thus, |C(a)|>=99 or 231 or 693. I know that 99 can't be it, since as you said C(a) contains an element of order 7. But, why can't it be 231?
    Also, A is a subgroup of C(a) because it has order p^2 for a prime p, and I have a theorem that says it is abelian as a result, so that conjugation is trivial.
     
  5. Dec 2, 2011 #4

    micromass

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    Because A is a subgroup of C(a). So |A| must divide |C(a)|.
     
  6. Dec 2, 2011 #5
    Oh wait, I messed up the first part. The two possibilities for the kernel are 1 and 7, and if it is 1 then |C/kernel|=7. Since the number of permutations of P is 9!, I'm unclear on why this leads to contradiction. Is there way to know |Aut(P)|<9! ?
     
  7. Dec 2, 2011 #6

    micromass

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    A is either [itex]\mathbb{Z}_9[/itex] or [itex]\mathbb{Z}_3\times \mathbb{Z}_3[/itex]. Can you calculate the automorphism groups of these two??
     
  8. Dec 2, 2011 #7
    Oh, I do have a theorem that the automorphism group of the cyclic group of order n is isomorphic to (Z/nZ)^x. So, since the co-prime elements of Z/9Z are 1,2,4,5,7,8 the automorphism group of Z9 has order 6. Likewise, the automorphism group pf Z3 has order 3. Does this mean that the automorphism group of Z3 x Z3 has order 9?
     
  9. Dec 2, 2011 #8

    micromass

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    That is correct.

    Sadly, no. Here is the trick to finding the order automorphism group of [itex]\mathbb{Z}_3\times \mathbb{Z}_3[/itex].

    • Show that each automorphism is also [itex]\mathbb{Z}_3[/itex]-linear. So our automorphism group is actually [itex]GL_2(\mathbb{Z}_3)[/itex].
    • An element in [itex]GL_2(\mathbb{Z}_3)[/itex] is determined by choosing a basis. How many possibilities are there to choose a basis of [itex]\mathbb{Z}_3\times \mathbb{Z}_3[/itex]?
     
  10. Dec 2, 2011 #9
    Just want to check that 0 is not an element of Z/9Z, right? Because then the order would be 7 and the proof would go out the window.

    What do you mean by Z3 linear? I guess if f is an automorphism, then f(x+y)=f(x)f(y). Not quite sure if this is what you mean...
     
  11. Dec 2, 2011 #10

    micromass

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    Well, [itex]\mathbb{Z}_3\times \mathbb{Z}_3[/itex] is a [itex]\mathbb{Z}_3[/itex]-vector space. So we can define a linear map as something that satisfies

    [tex]f(\lambda x+\mu y)=\lambda f(x)+\mu f(y)[/tex]
     
  12. Dec 2, 2011 #11
    Ok, so suppose f:P -> P is an automorphism. I guess I'm confused about how to go about showing that something is linear in Z3 if the function is defined on P, which has order 9.
     
  13. Dec 2, 2011 #12
    I guess if P=Z3 x Z3, then we'd have to define f((x,y))=(g(x), h(y)). Then, f is Z3 linear iff g and h are Z3 linear. Is this right?

    Also, I am talking about group isomorphisms, so I'm not sure how to deal with the constants.
     
  14. Dec 2, 2011 #13

    micromass

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    You must show for [itex]\lambda,x,y\in \mathbb{Z}_3[/itex] that

    [tex]f(\lambda(x,y))=\lambda f(x,y)[/tex]
     
  15. Dec 2, 2011 #14
    Ok, so just to recap, we want to find the number of automorphisms from Z3 x Z3 to Z3 x Z3. We want to show that f(x,y) is such an automorphism iff it is a linear transformation of Z3 x Z3 (represented by a 3x3 invertible matrix with entries from Z3). Ok, so if f(x,y) is an automorphism, and a,x,y are in Z3, then we need to show that af(x,y)=f(ax, ay). I am confused how to do this, because I only have theorems about addition.
     
  16. Dec 2, 2011 #15
    Oh, sorry a 2x2 invertible matrix
     
  17. Dec 2, 2011 #16

    micromass

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    Well, maybe your multiplication can be written as addition??
     
  18. Dec 2, 2011 #17
    ooh... so ax=x+x+..+x (a times), s0 f(x,y)+f(x,y)...+f(x,y) (a times)=f((x,y)+..+(x,y) (a times))=f(a(x,y)). Is that right? So then, af(x,y)+bf(z,w)=f(a(x,y))+f(b(z,w))=f(a(x,y)+b(z,w)). Thus, if f is an automorphism, it is a linear map in Z3xZ3. Do I also need to show that a linear map in Z3x Z3 is an automorphism of Z3xZ3?
     
  19. Dec 2, 2011 #18

    micromass

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    A bijective linear map will be an automorphism. You can show that if you like, but isn't that trivial??
     
  20. Dec 2, 2011 #19
    Also, I'm not sure how to find the number of bases of Z3 x Z3. It would have to consist of 2 linearly independent elements of Z3 x Z3.
     
  21. Dec 2, 2011 #20
    No, I do see why a bijective linear map is an automorphism. I know that the automorphism induces a linear map from Z3 x Z3 to Z3 x Z3, but do I know that it is bijective? Also, does a 2x2 matrix define a bijective linear map or just a linear map?
     
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