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December 9th, 2013, 08:58 AM   #1
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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.

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December 9th, 2013, 09:32 AM   #2
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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).
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December 31st, 2013, 06:06 AM   #3
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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 [GN(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.
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