5-Sylow Subgroup of Groups of Order 90.

[quantumdude]

Homework Statement

Show that the 5-Sylow subgroups of a group of order 90 is normal.

Homework Equations

None.

The Attempt at a Solution

I know that the number $\nu_5$ of 5-Sylow subgroups must divide 90 and be congruent to 1 mod 5. That means that $\nu_5\in\{1,6\}$. I also know that p-Sylow subgroups are normal iff they are unique, so it seems fairly obvious that I need to try to rule out $\nu_5=6$. I'm just not sure of how to do it. Can someone give me a nudge in the right direction?

 

[matt grime]

What else do we know? All Sylow-5 subgroups are conjugate. so there is a group hom from G to S_{v_5}. Dunno if that helps.

 

[quantumdude]

There's a group homomorphism between G and $S_{\nu_5}$ whether or not the 5-Sylow subgroups are conjugate. Or did you mean that there is a nontrivial homomorphism? If so, then I was not aware of that.

 

[matt grime]

Well, I'll repeat myself: all sylow subgroups are conjugate. This is part of hte Sylow theorems. Whether or not it helps here I don't know.

 

[quantumdude]

Yes, I know that Sylow II says that (by the numbering in my book, anyway). If I couldn't state the Sylow theorems, I'd be in big trouble! My point to you is that the fact that there is a group homomorphism from G to $S_{\nu_5}$ isn't dependent on Sylow II at all. There's obviously a homomorphism between any two groups. I was hoping that there was a more powerful constraint on that homomorphism that you left out of your response. But I guess there isn't.

 

[matt grime]

Since all sylow subgroups are conjugate, the image of the natural morphism is to a transitive subgroup of the permutation group. That is a strong condition, as ought to be clear, and was *my point to you*, so this is a highly non-trivial homomorphism. Plus one has the orbit stabiliser theorems, that I should not need to specify. But these are not guaranteed to be of any use. And if we act be conjugation on the sylow subgroups, then the sylow subgroups are in the (normal!) kernel of this map from the group to S_{v_d}

 

[quantumdude]

Since all sylow subgroups are conjugate, the image of the natural morphism is to a transitive subgroup of the permutation group. That is a strong condition, as ought to be clear, and was *my point to you*, so this is a highly non-trivial homomorphism.

Yes now that you've actually stated your point, it is clear that this is a highly nontrivial homomorphism. Thank you.

Plus one has the orbit stabiliser theorems, that I should not need to specify. But these are not guaranteed to be of any use.

The expression "orbit stabiliser theorem" doesn't appear in my book (neither does it appear there if I spell it "stabilizer"). Perhaps it goes by another name. I'll look into it.

And if we act be conjugation on the sylow subgroups, then the sylow subgroups are in the (normal!) kernel of this map from the group to S_{v_d}

I'll think more about this approach.

 

[matt grime]

Orbit Stabli(s|z)er:

If G is a group acting on a set X, then for all x in X |G|=|stab(x)||orb(x)|

it is one of the most fundamental results in group theory (it encodes Lagrange's theorem, etc, leads to the class equation, shows that the centre of a p-group is non-trivial). I'd think about getting a better book if it's your first course in group theory.

What does it mean here? Well, suppose that X is the set of cosets of a Sylow-5, then if there are 6 of them, and the action is transitive then the orbit has size 6. So the kernel stabili(s|z)er must have order 15. Dunno if that means much.

Actually scrub what I said above about the Sylow subgroups being in the kernel of the group hom - I must have had a glass of wine to many when I wrote that.

 

[quantumdude]

Orbit Stabli(s|z)er:

If G is a group acting on a set X, then for all x in X |G|=|stab(x)||orb(x)|

it is one of the most fundamental results in group theory (it encodes Lagrange's theorem, etc, leads to the class equation, shows that the centre of a p-group is non-trivial). I'd think about getting a better book if it's your first course in group theory.

OK, it's in my books. Only the term "stablizer" isn't used in the book for my course (Algebra, by Steinberger). It's called "isotropy group of x". But I've looked it up in Hungerford, and he states that "isotropy group of x", "subgroup fixing of x", and "stabilizer of x" are all synonymous.

I'm talking to my professor about this today. I'll compare notes with what you've written later. Thanks for your help.

 

[matt grime]

I think I solved this this morning, though I was sat in a hospital awaiting treatement at the time, so I don't guarantee my thinking. Suppose there are 6 Sylow-5 subgroups. Then the stabilizer under conjugation has order 15, so no element of the Sylow-4 subgroup can stabilize. But then this means wer have a group action of C_4 or C_2xC_2 on a set of size 6 with no points fixed by any non-trivial element, which doesn't happen (e.g. if Sylow-4=C_4=<g>, then orbits have length 2 or 4, and there is at least one orbit of length 2, thus g^2 has a fixed point, which we assumed couldn't happen since stab of any sylow-5 has order 15). A contradiction, hence there is only 1 Sylow-5 subgroup.

 

[quantumdude]

so no element of the Sylow-4 subgroup can stabilize.

What's a Sylow-4 subgroup? In all of the references I have, there is mention only of Sylow-p subgroups where p is prime.

 

[matt grime]

Sorry, I meant Sylow-2 (which has order 4).

 

[quantumdude]

No, the Sylow-2 subgroups of a group of order 90=2x3²x5 have order 2.

 

[matt grime]

Duh, I'm being stupid. Told you not to trust me.

 

[BSMSMSTMSPHD]

Wow, two Matt Grime mistakes in one thread - I never though I'd see the day.

 

[quantumdude]

I think I'm making progress, but I've still got quite a bit to do. My professor broke this problem down into pieces for us. He said to argue by contradiction, so assume that the number $\nu_5$ of 5-Sylow subgroups is equal to 6. He then sketched out the proof that this implies that there are 48 elements of order 15 and 24 elements of order 5. This is simple enough. But then he sketched out an argument that systematically rules out the various possibilities for elements of order 2, so that there is no 2-Sylow subgroup, thus contradicting Sylow I. It looks like a lot of brute force.

Here's an alternative method I've been thinking about.

1. Show that G has a subgroup H of order 45. Since [itex][G]=2[/itex], it follows that H is normal in G.
2. Show that H is characteristic in G.
3. Show that any 5-Sylow subgroup $P_5$ is contained in H.
4. Show that $P_5$ is characteristic in H.
5. Then, $P_5$ is normal in G by a Corollary from the book.
6. Then, $\nu_5=1$ by another Corollary from the book.

I decided to look at this route because I know that 1 is true (I just don't know how to prove it yet). What I want to know is, does anyone know off hand if 2, 3, or 4 are false? If so, then I won't waste my time with this.

Any advice is much appreciated.

 

[quantumdude]

More thoughts...

1. Show that G has a subgroup H of order 45. Since [itex][G]=2[/itex], it follows that H is normal in G.
2. Show that H is characteristic in G.

Still need help here.

3. Show that any 5-Sylow subgroup $P_5$ is contained in H.

H is solvable by the Feit-Thompson Theorem. Furthermore, since 45=5*9, and 5 is relatively prime to 9, Hall's Theorem guarantees that H has a subgroup of order 5. Furthermore, any two subgroups of H of order 5 are conjugate. I'm still not sure if this means that all of the 5-Sylow subgroups have to be contained in a subgroup of order 45.

4. Show that $P_5$ is characteristic in H.

Still not sure how to show this, but I'm looking into it.

As always, any advice is much appreciated.

 

[matt grime]

I don't know what it means to be 'characteristic', so I can't help there atm. However. If H is this group of order 45, then why are you invoking Hall's theorem to say it has a subgroup of order 5? It has one by Sylow, and it is unique, thus normal in H (though obviously, not necessarily in G). (Off the top of my head, the only theorem I can think of with Hall's name attached isn't even about group theory.)

 

[StatusX]

1. Show that G has a subgroup H of order 45. Since [itex][G]=2[/itex], it follows that H is normal in G.
2. Show that H is characteristic in G.
3. Show that any 5-Sylow subgroup $P_5$ is contained in H.
4. Show that $P_5$ is characteristic in H.
5. Then, $P_5$ is normal in G by a Corollary from the book.
6. Then, $\nu_5=1$ by another Corollary from the book.

You don't need 2, since a characteristic subgroup of a normal subgroup is normal, and for 4, any normal sylow p-subgroup is characteristic.

This might work. If there is such an H, then it's clear that n_5=1 for H. So it remains to see what happens if there is no such H.

 

[quantumdude]

Yes, showing that H exists is the key. I've been wracking my brain trying to do it, but I've hit a mental block. Besides, I don't think I should cite Feit-Thompson anyway, as I would have to cite a journal article (the proof is reportedly outside the scope of this course).

I think I might have more luck with the indirect approach.

So, $\nu_5\in\{1,6\}$, and the 5-Sylow subgroup $P_5$ is normal in $G$ iff it is unique.

Suppose $\nu_5=6$. Then, by a Corollary from my book, $[G:N_G(P_5)]=6$, the number of conjugates of $P_5$. Since $|G|=90$, this means that $|N_G(P_5)|=15$. Since $\mathbb{Z}_{15}$ is the only group of order 15, the normalizer of $P_5$ in $G$ must be cyclic.

Since there are 6 5-Sylow subgroups by hypothesis, let us index them: $P_5^{(i)}$, $i\in\{1,2,3,4,5,6\}$. My professor said to look at the intersection $N_G(P_5^{(i)}) \cap N_G(P_5^{(j)})$, $i \neq j$. He claims that we should find that there are no elements of order 15 in there, and thus no element of order 15 can normalize more than one 5-Sylow subgroup. This will enable us to count the number of elements of order 15. Also, since we know that the 5-Sylow subgroups of order 5 intersect only in the identity, we can cound the number of elements of order 5.

Elements of order 15: (8 generators of $\mathbb{Z}_{15}$)x(6 5-Sylow subgroups)=48 elements of order 15
Elements of order 5: (4 nonidentity elements of order 5)x(6 5-Sylow subgroups)=24 elements of order 5

So we've accounted for 72 of the 90 elements of $G$.

I just have to show that the intersection of the normalizers of distinct 5-Sylow subgroups can't contain an element of order 15. So if anyone can advise me on that, I'd be grateful. Then I can start looking at the 2-Sylow subgroups...

Sorry to switch my method of proof midstream like this. But this thing is due today, and I don't see myself coming up with the proof of the existence of the subgroup of order 45 by tonight.

 

[AndrewV]

I'm in this class with Tom and our other classmate Stan and I are pretty much at this same point in the proof. We're having a very difficult time figuring out how to find the subgroup of order 45. All we've found is the amount of elements in the 6 Sylow subgroups (24) and the amount of elements of order 15 (48)...

Tom, if you have any further information on the problem- go ahead and reply or private message me because Stan and I are at a standstill. Thanks.

 

[quantumdude]

Sure, I'll reply to this thread as I made progress.

 

[quantumdude]

OK, let's think about what it would take for an element of order 15 to show up in the intersection of the normalizers of two distinct 5-Sylow subgroups. $N_G(P_5^{(i)})$ is a subgroup of $G$ (I don't think Steinberger proved this, but it's easy to show), and it has order 15. It's a well known fact (don't remember where it is in the book) that the intersection of two subgroups is again a subgroup. So in order for an element (call it $x$) of order 15 to show up in the intersection, it would have to be in each of the normalizers to begin with. But that means that $N_G(P_5^{(i)})=N_G(P_5^{(j)})=<x>$, the subgroup of order 15 generated by $x$.

In other words, we need to show that two distinct 5-Sylow subgroups of $G$ can't have the same normalizer.

Agree so far?

 

[StatusX]

In other words, we need to show that two distinct 5-Sylow subgroups of $G$ can't have the same normalizer.

Right, and how many subgroups of order 5 can a group of order 15 have?

 

[quantumdude]

I count 4 of them:

<3>,<6>,<9>, and <12>. So there we go, they can't have the same normalizer and so there is no element of order 15 in the intersection of any two of them.

On to the 2-Sylow subgroups. More shortly.

 

[StatusX]

How about this. You've accounted for 72 elements. We know n_3=1 or 10, but since each 3-subgroup has 6 elements of order 9 (which can't be shared among different 3-subgroups), this forces n_3=1. Then let H be the unique, normal sylow 3-subgroup. Then look at G/H.

 

[StatusX]

I count 4 of them:

<3>,<6>,<9>, and <12>. So there we go, they can't have the same normalizer and so there is no element of order 15 in the intersection of any two of them.

You do realize these are all the same group right? If not, how do you think this give you what you want?

 

[quantumdude]

You're right, of course, there is only 1 subgroup of order 5. That was a mistake.

 

[quantumdude]

How about this. You've accounted for 72 elements.

I haven't really accounted for them yet, because I still haven't proven that the normalizers can't be the same.

We know n_3=1 or 10, but since each 3-subgroup has 6 elements of order 9 (which can't be shared among different 3-subgroups), this forces n_3=1. Then let H be the unique, normal sylow 3-subgroup.

I'm not clear as to why there must be 6 elements of order 9 in each 3-Sylow subgroup, or why they can't be shared. Could you explain that?

Then look at G/H.

Will do, as soon as I get the above nailed down.

 

[quantumdude]

I haven't really accounted for them yet, because I still haven't proven that the normalizers can't be the same.

Oh wait, sure I have. For any subgroup $H$ of $G$, $H$ is a normal subgroup of its own normalizer. So 6 distinct 5-Sylow subgroups of $G$ can't have the same normalizer, because there isn't enough room in there for them. (Actually, this is what I was thinking when I mistakenly thought there were 4 subgroups of order 5 in a group of order 15).

 

[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.