5-Sylow Subgroup of Groups of Order 90.

[StatusX]

Sorry, I was assuming the 3-subgroups are cyclic. If they are, then there are 6 numbers in 0,...,8 relatively prime to 9, so 6 elements in Z_9 with order 9, and since these generate the entire subgroup they can't be in any proper subgroup (like a non-trivial intersection with another 3-subgroup).

If the subgroups aren't cyclic, they're Z_3 x Z_3. Any 2 of them intersect in at most 3 elements (the largest non-trivial subgroup of Z_3 x Z_3) including the identity. So two of the subgroups account for at least 14 distinct non-identity elements. A third can share at most 3 with each of these, including the identity, so this adds another 4, and you're already at the 18 elements you have left with no room for, say, the identity (among other things).

 

[quantumdude]

Yeah, our professor explicitly told us to avoid looking at the 3-Sylow subgroups. He sketched out an argument for ruling out the possibilities for the number of 2-Sylow subgroups one by one.

I'm working on fleshing that out now. I'll post again once I've made some progress.

Thanks to everyone for all your help.

 

[StatusX]

Yeah, our professor explicitly told us to avoid looking at the 3-Sylow subgroups.

What? Why?

 

[quantumdude]

Because the 2-Sylow subgroups intersect only in the identity, and so they're much cleaner. And because he is very familiar with this material, he knows already that we will be able to rule out all possibilities for the number \nu_2 of 2-Sylow subgroups, thus contradicting the first Sylow Theorem.

 

[StatusX]

I guess, but I just laid out the whole argument for you. If you can show n_3=1 (which, admittedly, isn't the cleanest thing in the world) then you have |G/P_3|=10, so its 5-subgroup corresponds by the lattice isomorphism theorem to a subgroup of G of order 45, and you're done. That's not to say you shouldn't try to find another way to do it, though.

 

[quantumdude]

I guess, but I just laid out the whole argument for you.

Ah, so you have! I replied to you before I read your edit.

Since I still have time I think I'm going to try to do it the way my professor suggested as well. One of his favorite sayings throughout the entire semester has been that it's good to be able to prove things more than one way.

OK, here we go. We know that G has 1 element of order 1, 24 elements of order 5, and 48 elements of order 15. That accounts for 73 elements of G.

The number \nu_2 of 2-Sylow subgroups are:

\nu_2\in\{1,3,5,9,15,45\}

Because the 2-Sylow subgroups intersect only in the identity, there is a 1-to-1 correspondence between 2-Sylow subgroups and elements of order 2. S we can rule out \nu_2=45 immediately.

Next up: \nu_2=15.

 

[AndrewV]

Since I still have time I think I'm going to try to do it the way my professor suggested as well. One of his favorite sayings throughout the entire semester has been that it's good to be able to prove things more than one way.

So does this mean you've found a way to solve the proof? If so, do you think you could give some more info to me? I can't get any further than what I stated in my other post.

 

[quantumdude]

So does this mean you've found a way to solve the proof?

No, StatusX just gave it to me, in Posts 31 and 35. The only tricky parts are:

1. Proving that, if two distinct 3-Sylow subgroups P_3 and Q_3 are isomorphic to \mathbb{Z}_3\times\mathbb{Z}_3 then they intersect in at most 3 elements. That's what I'm working on now.

2. Finding and proving the so-called Lattice Isomorphism Theorem (aka the Fourth Isomorphism Theorem). This wasn't covered in class, and it isn't in the book. So a-hunting I will go.

 

[AndrewV]

What do you think's going to happen if none of us turn in the problem? At this rate, I don't see us really getting close.

 

[quantumdude]

I fully expect to get done tonight. StatusX has reduced this to 2 tasks, as I've listed. And finding the proof of the Fourth Isomorphism Theorem somewhere online can't be difficult. Locating proofs of famous theorems hardly ever requires much work.

 

[AndrewV]

Okay sounds awesome, I'm going to go ahead and go through StatusX's proof and make sense of it and write it down - if I run into any more issues, I'll be sure to post again. :)

 

[AndrewV]

Ah, so you have! I replied to you before I read your edit.

Since I still have time I think I'm going to try to do it the way my professor suggested as well. One of his favorite sayings throughout the entire semester has been that it's good to be able to prove things more than one way.

OK, here we go. We know that G has 1 element of order 1, 24 elements of order 5, and 48 elements of order 15. That accounts for 73 elements of G.

The number \nu_2 of 2-Sylow subgroups are:

\nu_2\in\{1,3,5,9,15,45\}

Because the 2-Sylow subgroups intersect only in the identity, there is a 1-to-1 correspondence between 2-Sylow subgroups and elements of order 2. S we can rule out \nu_2=45 immediately.

Next up: \nu_2=15.

I think right about now this is the method I'm going to try to solve the problem. Thanks everyone for their help so far. Even if I'm not going to use a method someone posted, it's really nice to see how people would solve it, and the devotion people have to helping everyone else.

 

[StatusX]

Yea, sorry, I didn't mean to do someone's HW for them. Since this thread has been here for a while, and since other people had tried to give out solutions, I assumed whatever deadline had passed. I'd feel better if you tried to find your own way to do it, and I think looking at the 2-subgroups as you've mentioned is a good way.

 

[AndrewV]

Yea, sorry, I didn't mean to do someone's HW for them. Since this thread has been here for a while, and since other people had tried to give out solutions, I assumed whatever deadline had passed. I'd feel better if you tried to find your own way to do it, and I think looking at the 2-subgroups as you've mentioned is a good way.

Our Professor told us to "ask anyone" about the question or for any help. I'm going to look at the 2-subgroups...I don't think I'll personally be able to figure it out, but I do appreciate your effort, StatusX, helping us with the problem.

 

[quantumdude]

Yea, sorry, I didn't mean to do someone's HW for them.

Don't feel bad. I've been working it out your way, and there are still plenty of steps to fill in. Our professor gave us the same level of help for the other approach (ie, the approach that considers the number of 2-Sylow subgroups).

And if it makes you feel any better, I've figured out a way to show the existence of a group of order 45 without using the lattice isomorphism theorem, so I'm not using that suggestion at all. (Thanks for the tip though, it gives me something to study over the summer).

Thanks for everything.

 

[AndrewV]

Tom, do you think we could see your proof to show the existence of a group of order 45, because Steinberger was telling us that it's a really important thing to prove.

 

[quantumdude]

Sorry, I meant that I figured out how to prove that a subgroup of order 45 exists under the assumption that P_3 is normal in G. I'm not at the end just yet...

 

[AndrewV]

It's totally cool, Tom. Does anyone think they can give me a push in the right direction on how to find the order of the 2-Sylow Subgroup? I understand what Tom said about why it can't be 45, but I'm not sure how to figure out the other ones.

If \nu_{2} = 15 then there would be fifteen elements of order 2 (not including the identity) which would give us (72+15 = 87 + identity = 88 classified elements) so there is now only room for 2 more elements from the 3-Sylow subgroup. So is that why it can't be 15? Let me know if I'm just spewing out random things and not remembering how to do things right.

My problem is that I was always a chapter behind in this class so I was finally just starting to understand SemiDirect Products and Extensions.

 

[quantumdude]

I did in fact do this by looking at the 3-Sylow subgroups. So I can't tell you exactly how to rule out all the possibilities for \nu_2. But here are the clues that Steinberger gave us in Friday's help session.

\nu_2=15 and \nu_2=45 were ruled out because, he claims, G can have at most 10 elements of order 2. I'm not sure of why he claimed that though.

\nu_2=3 and \nu_2=9 were ruled out because they lead to, respectively, |N_G(P_2)|=30 and |N_G(P_2)|=10, which are both divisible by 5. To complete the argument, he said to think about N_G(P_5), and he asked if the 5-Sylow subgroup of any group of order 10 or 30 is normal. Well, in response to that I say that the 5-Sylow subgroup of any group of order 10 is obviously normal, as it has index 2. And he proved that the 5-Sylow subgroup of any group of order 30 is normal in Lemma 5.3.19. So the answer to his question is "yes". So we see that 5-Sylow subgroups are normalized by groups of orders 10 and 30, contradicting our previous assumption that the normalizer of P_5 has order 15. At least I understood that one!

In Monday's help session I asked him how he ruled out \nu_2=1. He said that in this case P_2 is not only normal, but also central. So it's possible that there are elements of order 6 and 10. He claims this is not possible. I haven't taken the time to figure out why.

That leaves only one case: \nu_2=5. He left it to us to derive a contradiction from this, as well as filling in the details of the proofs he sketched out above.

Do all that, and you're done.