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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

```

```
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/
```
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