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Sylow first thm and normal subgroups

  1. Feb 28, 2006 #1
    I just finished learning about Sylow theorems and I have a quick question and was wondering if someone could help me see something. In this handout I have from a section on Sylow it gives a lemma to the first theorem and it says;

    Suppose p is prime and |G| = p^n then G has normal subgroups of order p^k for all k between 1 and n.

    Then it says that this is obvious from Sylow 1st thm. Obvious? How? Sylow 1st thm doesnt mention anything about the subgroups being normal. I would like to understand this before moving on.

  2. jcsd
  3. Mar 1, 2006 #2


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    Yeah, I definitely don't see how it's obvious. My book gives the first Sylow theorem as

    If G is a finite group, and pm is the highest power of p dividing |G|, then G contains a subroup of order pm

    Another one of the theorems says that all such subgroups are conjugate, so a corollary is that if G has only one Sylow-p subgroup, it is normal. However, the lemma you're referring to doesn't seem obvious, especially not from the Sylow theorems, and even more especially not from the first one.

    I don't know how to prove it, but I might guess using the Orbit-Stabilizer theorem, where G acts on subgroups of G by conjugation?
  4. Mar 1, 2006 #3
    Well, I don't really see how it's related to Sylow's theorems. They can't really say much about p groups...

    The proof, however, is not too difficult:
    It's clearly true for n=0. Assume it's true for k<n, and look at Z(P). It is known (and if it's not to you, just ask and I'll write the proof) that Z(P) where P is a p groups is non-trivial. P/Z(P) is therefore a p group of order p^k, k<n, and by the induction assumption has a normal subgroups of any order p^q 1<q<k (including 1 and k). Any such group, N', is clearly the image of a group containing Z(P), i.e. N'=N/Z(P).

    N is of order p^(q+n-k). Let's show that it is normal in P. We know, for all g in P, that
    \forall n \in N\exists n' \in N:\left[ {gZ\left( P \right)} \right]\left[ {nZ\left( P \right)} \right]\left[ {g^{ - 1} Z\left( p \right)} \right] = n'Z\left( P \right) \\
    \Rightarrow gng^{ - 1} n'^{ - 1} = x \in Z\left( P \right) \subset N \Rightarrow gng^{ - 1} = xn' \in N \Rightarrow gNg^{ - 1} \subset N \\
    \Rightarrow gNg^{ - 1} = N \\

    And we've found a normal subgroup of any order p^l, for n-k+1<l<n.
    All we need to do now is find such a subgroup for 1<l<n-k (Note that I'm including the edges). But any subgroup of Z(P) is clearly normal in P, and since it is abelian, it has a subgroup of any order dividing its order, i.e. of any order P^l s.t. 1<l<n-k!
    That proves the lemma.
  5. Mar 1, 2006 #4


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    How do you know this?
    I guess you mean that, in addition, N is such that N/Z(P) is normal in P/Z(P).
  6. Mar 1, 2006 #5
    1. This is a fundamental fact, isn't it? There is a one to one correspondence between subgroups og G/A and subgroups of G containing A. Anyway, if f denotes the natural projection, then N is given simply by f^-1(N').

    2. N is such that N'=N/Z(P). N' was taken as the promised normal subgroup in P/Z(P).
  7. Mar 1, 2006 #6

    matt grime

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    It would appear that if you'd googled for 'sylow theorem' you'd've got your answer straigh away:

    Theorem 3 (Sylow's First Theorem)
    Let |G| = p^nr, where r is not divisible by p. Then for each 0 <=i <= n-1, every subgroup of G of order p^i is contained in a subgroup of order p^i+1, and is normal in it. In particular, every p-subgroup of G is contained in a Sylow p-subgroup of G.
  8. Mar 1, 2006 #7
    Now I feel stupid because I don't see how this proves the Lemma.:confused:

    Could you explain?

    By the way, is there anything wrong with my proof?
  9. Mar 1, 2006 #8


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    matt grime

    That doesn't appear to prove that G of order pn has normal subgroups of order pi for all i, 0 < i < n+1. In fact, it doesn't even prove the existence of subgroups (normal or otherwise) of order pi for all i.


    Yeah, sorry, it seems I'm rusty with this stuff (which isn't good, because it's only been a year). That proof looks fine.
  10. Mar 2, 2006 #9

    matt grime

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    Sorry, AKG, I must disagree: it seems rather trivial to demonstrate te existence of subgroups of the correct order from that copy and paste of mine.

    Firstly, it is trivial that G has a subgroup of order p, now apply Sylow's first theorem as given in my copy and pasted link to obtain a normal series of subroups of all orders.

    I can not recall, nor can i be bothered to prove if A normal in B and B normal in C implies A normal in C, however this seems to be the only point you need to think about.
    Last edited: Mar 2, 2006
  11. Mar 2, 2006 #10


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    In general this is false, a couple of counter examples here:

    http://planetmath.org/encyclopedia/NormalityOfSubgroupsIsNotTransitive.html [Broken]

    For the OP- they may mean it's "obvious" from the proof you were given of Sylow's 1st theorem. In this specific case of a prime power you may be able to get more information from the proof.
    Last edited by a moderator: May 2, 2017
  12. Mar 2, 2006 #11

    matt grime

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    I had the feeling that it was false in general, however, in this case we have the extra information of a complete tower of (maximal) subgroups to use too: we can also work from the top down quotienting out by the maximal normal subgroup, and perhaps applying induction in this maximal subgroup, ie given normal H<G of orders p^{n-1} and p^n respectively we may assume that H has a complete subset of normal subgroups of all orders.
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