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 September 17th, 2009, 09:05 PM #1 Member   Joined: Aug 2008 Posts: 84 Thanks: 0 Normal sylow subgroup of a group of order p^aq^b So this is a curious little question, where p and q are of course primes. Not quite sure how to go about it. Clearly the case p = q is of no interest. I think my current repertoire of techniques in group theory isn't enough to handle this problem. Perhaps sections one and two of dummit and foote chapter 6 might shed some light. Appreciate any helpful hints, this problem has been annoying me.
 September 19th, 2009, 07:02 PM #2 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Normal sylow subgroup of a group of order p^aq^b It can be shown that Sylow p-subgroup is normal if and only if it is the only Sylow p-subgroup. It requires that $b$ is not multiple of $p-1$ and $a$ is not multiple of $q-1$ In this case our group is a direct product of two Sylow groups.
 September 20th, 2009, 12:03 PM #3 Member   Joined: Aug 2008 Posts: 84 Thanks: 0 Re: Normal sylow subgroup of a group of order p^aq^b So you're saying that we must impose the conditions that b is not a multiple of p-1 and a is not a multiple of q-1 for this to be true? The original question is verbatim from a qualifying exam.
 September 21st, 2009, 10:53 AM #4 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Normal sylow subgroup of a group of order p^aq^b Conditions are coming from the fact that it has to be only one p-Sylow subgroup and only one q-Sylow subgroup. You can check Sylow theorem II, first and second statements of it. P.S. I failed number qualifying exams in the past.
 September 21st, 2009, 01:53 PM #5 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Normal sylow subgroup of a group of order p^aq^b How does (q-1)|a relate to the uniqueness of a Sylow-p subgroup? There's probably some blindingly obvious number theory I'm missing, but I don't see the relation between the 2 ideas.
 September 21st, 2009, 05:06 PM #6 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: Normal sylow subgroup of a group of order p^aq^b Here are full path: Sylow theorem states that if G is finite group and $|G|= p^am$ with m not divisible by $p$, then: (a) Any two Sylow p-subgroups are conjugate. (b) Let $s_p$ be the number of Sylow p-subgroups in G; then $s_p= 1 (mod(p))$ and $s_p$ divides $m$ (c) Every p-subgroup of G is contained in a Sylow p-subgroup To be the only subgroup, $s_p$ must be $= 1$, in our case $m= q^b$, so $s_p= q^c, c <= b$. Order of $q$ in $Z_p$ is $p-1$, so (here is my correction) if $b=>= p-1$, then exists c so that $s_p$ may not be 1. Similar thoughts about q. Overall $b < p-1, a < q-1$ Now If N is the only subgroup, it is clearly normal (actually, it is also a characteristic subgroup). Otherwise, if N is normal (a) of Sylow theorem states that is the only one.
September 21st, 2009, 09:52 PM   #7
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Re: Normal sylow subgroup of a group of order p^aq^b

Quote:
 Originally Posted by zolden Order of $q$ in $Z_p$ is $p-1$, so (here is my correction) if $b=>= p-1$, then exists c so that $s_p$ may not be 1.
Ah! That is what I was missing. I'm not completely sure I understand the last step, but I need to sit down and work it out-- I "believe" you, now, though.

(Sorry for hijacking this thread)

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