Applying the Sylow Theorems

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Summary:

I'm trying to get a little bit more comfortable with using the Sylow theorems by reworking examples from my abstract algebra textbook.

Main Question or Discussion Point

I'm trying to get a little bit more comfortable with using the Sylow theorems and looking back at some examples from my abstract algebra textbook (we used Fraleigh). In an example involving groups of order 36 (the claim is no such groups are simple), the author writes:

"Such a group ##G## has either one or four subgroups of order 9. If there is only one such group, it is normal in ##G##. If there are four such subgroups, let ##H## and ##K## be two of them. As an in Example 4.3.13, ##H \cap K## must have 3 elements, or ##HK## would, have 81 elements which is impossible. Thus the normalizer of ##H \cap K## has as order a multiple of >1 of 9 and a divisor of 36; hence the order must be either 18 or 36. If the order is 18, the normalizer is then of index 2 and therefore is normal in ##G##. If the order is 36, then ##H \cap K## is normal in ##G##."

I'm trying to rework this example from a different angle and I'm wondering if we could alternatively look at the subgroups of order 4 for which ##G## would have either 1 or 9. If it has 1 then it's normal and we're done. If it has 9, then taking a similar approach, let ##H## and ##K## be two such subgroups. Then if the intersection of ##H## and ##K## is trivial, then ##G## has 27 elements of order 4, leaving us with 8 elements whose order must divide 36 (so either 6, 9, 12, 18, or 36). We know that we have at least 1 Sylow 3-subgroup of order 9 so the remaining elements must be contained in this subgroup. So in fact it's the only Sylow 3-subgroup, so it's normal and we're done. OTOH if ##H \cap K## is nontrivial, then since it's order is a multiple of 4 (greater than 1), and divides 36, it has to have an order of 12 or 36. But both of these are impossible since ##H \cap K## cannot have an order greater than 4. So ##H \cap K## would have to be trivial.
 
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  • #2
Infrared
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Then if the intersection of ##H## and ##K## is trivial, then ##G## has 27 elements of order 4
I don't think this is right. It looks like you're assuming that any two of the ##2##-Sylow subgroups intersect trivially, and adding up all ##9\times 3## non-identity elements from each subgroup gives you ##27## total. However, you only assumed that the two subgroups ##H## and ##K## intersected trivially, not that all ##2##-Sylow subgroups did. It's possible for some of them to intersect trivially, and for others to have a nontrivial intersection.

Also, these elements don't have to have order ##4##- they could have order ##2## instead (not that this matters to your argument)
 
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  • #3
I don't think this is right. It looks like you're assuming that any two of the ##2##-Sylow subgroups intersect trivially, and adding up all ##9\times 3## non-identity elements from each subgroup gives you ##27## total. However, you only assumed that the two subgroups ##H## and ##K## intersected trivially, not that all ##2##-Sylow subgroups did. It's possible for some of them to intersect trivially, and for others to have a nontrivial intersection.

Also, these elements don't have to have order ##4##- they could have order ##2## instead (not that this matters to your argument)
Ah, okay. I see what you're saying. Could it be reworked by arguing that a nontrivial intersection of any two Sylow subgroups has exactly order 2? Therefore the elements would have to be of order 2. So the number of distinct elements belonging to the 9 Sylow 2-subgroups is either 9 or 27. But if it's 9 then there are 26 remaining elements, 9 of which must have an order of 9 or 3 since ##G## has at least 1 Sylow 3-subgroup. But then there aren't enough elements remaining to construct three more Sylow-3 subgroups, so there would only be one.
 
  • #4
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a nontrivial intersection of any two Sylow subgroups has exactly order 2
This is true, but you still don't know how many of the intersections are trivial, and how many have size two. So I don't know how you conclude

So the number of distinct elements belonging to the 9 Sylow 2-subgroups is either 9 or 27
There are lots of different ways the ##2##-Sylow subgroups could intersect. Let's say the subgroups of order ##4## are ##H_1,\ldots,H_9##. Then maybe ##|H_1\cap H_2|=1## and ##|H_2\cap H_3|=2##, etc. There are many possibilities for how big their union is, not just ##9## or ##27##.

I'm also not sure how you're getting a 9 element union as a possibility.
 
  • #5
This is true, but you still don't know how many of the intersections are trivial, and how many have size two. So I don't know how you conclude



There are lots of different ways the ##2##-Sylow subgroups could intersect. Let's say the subgroups of order ##4## are ##H_1,\ldots,H_9##. Then maybe ##|H_1\cap H_2|=1## and ##|H_2\cap H_3|=2##, etc. There are many possibilities for how big their union is, not just ##9## or ##27##.

I'm also not sure how you're getting a 9 element union as a possibility.
Ah I think I got a head of myself, those were merely best and worst case scenarios (if all intersections were nontrivial and trivial, respectively); I keep forgetting that there are multiple ways for these groups to intersect. Alright, I'm not sure I can salvage the argument. Back to the drawing board. Thank you for taking the time to respond, I greatly appreciate it.
 
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  • #6
mathwonk
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SPOILER ALERT:

If you want a hint:

The case of #(G) = 36, is one of the "easy" cases to prove G not simple. In particular there is no need to examine the intersections of the various subgroups. It follows quickly just from the fact that 36 < 4! = 24.

You can just use the basic principle that any group permutes its sylow p-subgroups transitively by the action of conjugation.

This technique is explained in detail in the notes for my algebra course math 843-1, on my insecure webpage at UGA. It is shown there that all groups of order < 168 which are not of prime order, and not isomorphic to A(5), must have a non trivial normal subgroup.

The point is that for some reason the author of your book is making a trivial example look hard. Maybe he wants to give you hands on experience playing with elements of groups, but he is not showing you the more powerful way to do this problem, which I recommend understanding.

I.e. a grup is non simple if and only if it has a homomorphism to another group, which is not constant and not injective. In particular this holds if the map is not constant and the target group has order not a multiple of the order of your group. Then use the fact that for any subgroup of your group, your group permutes both its cosets, by translation, and its conjugates, by conjugation. This defines homomorphisms into various permutation groups. In particular, it follows that if your group has order n, and has exactly r sylow p groups, where n does not divide r!, then your group is not simple.

The message is that using mappings is often more effective than using elements.

(not secure):
http://alpha.math.uga.edu/~roy/
 
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