# 5-Sylow Subgroup of Groups of Order 90.

1. May 9, 2007

### Tom Mattson

Staff Emeritus
1. The problem statement, all variables and given/known data
Show that the 5-Sylow subgroups of a group of order 90 is normal.

2. Relevant equations
None.

3. 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?

2. May 9, 2007

### 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.

3. May 9, 2007

### Tom Mattson

Staff Emeritus
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.

4. May 9, 2007

### 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.

5. May 9, 2007

### Tom Mattson

Staff Emeritus
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.

6. May 9, 2007

### 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}

Last edited: May 9, 2007
7. May 9, 2007

### Tom Mattson

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

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.

Last edited: May 9, 2007
8. May 10, 2007

### 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.

9. May 11, 2007

### Tom Mattson

Staff Emeritus
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.

10. May 11, 2007

### 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.

Last edited: May 11, 2007
11. May 13, 2007

### Tom Mattson

Staff Emeritus
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.

12. May 13, 2007

### matt grime

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

13. May 14, 2007

### Tom Mattson

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

14. May 14, 2007

### matt grime

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

15. May 14, 2007

### BSMSMSTMSPHD

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

16. May 14, 2007

### Tom Mattson

Staff Emeritus
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 $[G]=2$, 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.

Last edited: May 14, 2007
17. May 14, 2007

### Tom Mattson

Staff Emeritus
More thoughts...

Still need help here.

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.

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

As always, any advice is much appreciated.

18. May 14, 2007

### 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.)

19. May 15, 2007

### StatusX

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.

Last edited: May 15, 2007
20. May 15, 2007

### Tom Mattson

Staff Emeritus
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.