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Proof of Normal Sylow p-Subgroup

Date: 10/28/2005 at 16:52:26
From: Asif
Subject: Sylow p-subgroup 

Let G be a finite group in which (ab)^p = a^p*b^p for every a,b in G,
where p is a prime dividing o(G).  Prove:

  a. the Sylow p-subgroup of G is normal
  b. if P is Sylow p-subgroup of G, then there exists a normal
     subgroup N of G with P intersection N is identity and PN = G.
  c. G has a nontrivial center.

The most confusing thing is I don't know the order of Group G.  First
I thought that I can show that G is Abelian and then part a and c
are immediate consequences of Abelian group, because all subgroups
are normal in Abelian and the center is entire group.  If p = 2, then
ab = ba but if p > 2, then I can not show that ab = ba.

My second thought was to show that there exists only one Sylow 
p-subgroup which implies that the subgroup is normal.  But I don't 
know the order of G in order to apply Sylow theorem part 3 to find #p
subgroup.  Can you please help me?



Date: 10/29/2005 at 08:08:12
From: Doctor Jacques
Subject: Re: Sylow p-subgroup

Hi Asif,

The hypothesis tells us that the function f : G -> G defined by
f(g) = g^p is an endomorphism (a homomorphism of G into itself).

For any positive integer k, let us denote by f_k the k-iterate of f, 
i.e.:

  f_k = f o f ... o f (k times)

f_k is the composition of homomormhisms, and is therefore also a 
homomorphism.  Note that we have:

  f_k(x) = x^(p^k)          [1]

and the homomorphism property tells us that, for any a,b in G, we 
have:

  (ab)^p_k = a^(p^k) * b^(p^k)       [2]

(this can also be shown directly by induction).

The kernel K of f_k is the set {x in G | x^(p^k) = 1}; this is a 
normal subgroup of G.

As any element of K has order p^i for some i, K has order p^j for 
some j (if the order of K were divisible by another prime q, then K 
would contain an element of order q)--in other words, K is a p-group.

Any p-group is contained in some Sylow p-subgroup of G.  As K is 
normal, it is contained in the intersection of all Sylow p-subgroups 
of G, the so-called p-core of G (do you see why?).

Let us now choose k such that p^k is the highest power of p that 
divides |G|--the Sylow p-subgroups of G have order p^k.

If P is any Sylow p-subgroup of G, then P has order p^k, and any 
element x of P satisfies x^(p^k) = 1, by Lagrange's theorem.  This 
means that P is a subgroup of K; as we already have seen that K is a 
subgroup of P, we have K = P, and this shows that P is normal in G, 
which was question (1).

To summarize, for that fixed value of k, we have a homomorphism:

  f_k : G -> G : f_k(x) = x^(p^k))     [3]

whose kernel is P.  Let N = f_k(G) be the image of f_k; this is a 
subgroup of G.

N is isomorphic to the quotient group G/P, and its order |N| = |G|/p^k
is relatively prime with p, by the definition of a Sylow p-subgroup.
This implies that the intersection of N and P is trivial, since any
element in the intersection has order dividing p^k and |N|.

We want to show that G is the internal direct product P x N - this 
will answer question (2).

Note that the elements of N are representatives of the distinct 
cosets of P in G.  Indeed, if two elements x and y are in the same 
P-coset, the element xy^(-1) of N is in P; as the intersection of P 
and N is {1}, this implies x = y.  As |N| = [G:P], this shows that N 
contains exactly one element of each coset of P.

As any element g of G lies in some coset of P, we have therefore g = 
hn, for some n in N and h in P; in other words, G = PN.  Note that the 
expression g = hn is unique, since n is uniquely determined by the 
P-coset it belongs to.

Let us take elements h in P and n in N.  Because h is in the image of 
f_k, we have h = x^(p^k) for some x in G.  We have:

  h * n * h^(-1) = h * x^(p^k) * h^(-1)
                 = (h * x * h^(-1))^(p^k)          [4]
                 = h^(p^k) * x^(p^k) * h^(-p^k)    [5]
                 = x^(p^k)                         [6]
                 = n                               [7]

In these equations, [4] is true in any group, because conjugation is 
an automorphism.  [5] follows from the hypothesis on G (easily 
generalized by induction to the product of more than 2 factors).  [6] 
results from the fact that h is an element of P, and |P| = p^k, and 
[7] results from the definition of x.

[7] can also be written as:

  hn = nh                                           [8]

To summarize, we have two subgroups N and P such that:

* Each element of N commutes with each element of P ([8])

* Each element g of G is uniquely expressible in the form g = hn,
  with h in P and n in N

These two facts allow us to conclude that G is isomorphic to the 
direct product P x N.  As each direct factor is normal, this shows 
that N is normal in G (this was question 2).

For question 3, you can use the theorem that says that any p-group 
has a non-trivial center.  If Z(P) is the center of P, then Z(P) x 
{1_N} is contained in the center of G = P x N.

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 



Date: 10/29/2005 at 12:11:42
From: Asif
Subject: Thank you (Sylow p-subgroup )

Thank you very much for your explanation and reply.  I have a question
on the first part of the problem.  Would this approach work:

Suppose a,b belongs to Sylow p-subgroup.  e is the identity element of
the same subgroup.  It is given Sylow p-group has order p, because p
divides O(G).  Therefore 

  a^p = e and b^p = e 

and

  e = (ab)^p = a^p.b^p = e.e = b^p.a^p = (ba)^p

then ab = ba.  Therefore Sylow p-subgroup is Abelian, therefore it is
normal.  Is that correct?



Date: 10/30/2005 at 05:02:33
From: Doctor Jacques
Subject: Re: p-Sylow subgroup

Hi again Asif,

There are a few flaws in your argument.

First, the order of a Sylow p-subgroup is not necessarily p, it is 
p^k, the largest power of p that divides the order of the group (this 
is the definition of a Sylow subgroup).  For example, in S_4 (of order 
24), the Sylow 2-subgroups have order 8.  If a Sylow p-subgroup has 
order p, we do not need any special hypothesis--any group of prime 
order is cyclic and therefore Abelian.

Second, even assuming that we replace p by p^k (this is legitimate, 
as explained in my previous answer), all you manage to prove is that:

  (ab)^(p^k) = (ba)^(p^k) (= e)

but this does not imply that ab = ba.  In any group of order n, we 
have (ab)^n = (ba)^n = e, but this does not imply that all groups are 
Abelian...

Third, and this is the most critical point, the fact that a subgroup 
is normal has nothing to do with the fact that it is Abelian--these 
are two completely different concepts.

Being Abelian is an intrinsic property of a group--it can be checked 
by using only the elements of the group.

On the other hand, being normal is a relation between a subgroup and 
a group containing it--a group is not "normal" by itself, it can be 
a normal subgroup of a group.  The correct formulation is "H is normal 
in G", when H is a subgroup of G (it is true that the "... in G" part 
is often omitted, when it is clear from the context).

For example, any group is a normal subgroup of itself, even when it 
is not Abelian.

For an example in the other direction, in S_3, the subgroup {(),(1,2)}
is Abelian (it is cyclic of order 2), but it is not a normal subgroup
of S3.

I hope this helps clarify a few concepts--please feel free to write
back if you want to discuss this further.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Modern Algebra

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