Proof of Normal Sylow p-SubgroupDate: 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/ |
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