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 December 9th, 2013, 08:58 AM #1 Senior Member   Joined: Feb 2013 Posts: 153 Thanks: 0 Sylow subgroup Hi. I dont know how to solve following problem: Let G be the group of order p*q^2, where p and q are different prime numbers. Prove that at least one Sylow subgroup of group G is normal. Thanks. December 9th, 2013, 09:32 AM #2 Math Team   Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: Sylow subgroup I'd approach it by noting that the number of q-Sylow subgroups of G must be 1, p or p^2 and analyzing each case. I am not pretty good at Sylow groups, so I'll leave it to you (or anyone else that might want to show an approach. Note that the above is not a hint as I haven't calculated up anything yet. In fact, the approach might not even work). December 31st, 2013, 06:06 AM #3 Senior Member   Joined: Mar 2012 Posts: 294 Thanks: 88 Re: Sylow subgroup I will assume that you mean |G| = pq^2, and not (pq)^2. We will consider 3 cases. We will use P to denote a Sylow p-subgroup, and Q to denote a Sylow q-subgroup. It is clear that P has 1,q or q^2 conjugates, and Q has either 1 or p conjuagtes. Case 1: p not congruent to 1 (mod q). In this case we have [G:N(Q)] = 1 (mod q), so [G N(Q)] cannot be p. Thus Q is normal in G. Case 2: p = 1 (mod q), q^2 not congruent to 1 (mod p). This implies q is not congruent to 1 (mod p), for if so: q = kp + 1, so q^2 = (kp + 1)^2 = k^2p^2 + 2kp + 1, a contradiction. Since we have [G:N(P)] = 1 (mod p), this means that [G:N(P)] is not q or q^2, hence P is normal in G. Case 3: p = 1 (mod q), q^2 = 1 (mod p). Here, q^2 = kp + 1 = k(mq + 1) + 1 = (km)q + 2, so 2 = q(q - km). This implies q|2, so q = 2. Hence p = 3, so we have a group of order 12. If P (of order 3) is not normal in G, we must have 4 Sylow 3-subgroups, since 2 is not congruent to 1 (mod 3). This gives us 8 elements of order 3, leaving just 4 elements left over, which must then be the sole Sylow 2-subgroup of order 4. Tags subgroup, sylow Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Colocha07 Abstract Algebra 0 May 23rd, 2010 04:52 PM envision Abstract Algebra 1 October 4th, 2009 03:24 AM Spartan Math Abstract Algebra 6 September 21st, 2009 09:52 PM Erdos32212 Abstract Algebra 0 December 8th, 2008 06:15 PM weier Abstract Algebra 1 November 20th, 2006 10:40 AM

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