in linair programming

that suggests that x j z j = 0, for j = 1, 2, . . . , n, suggests that either xj or zj is 0 right?

That seems weird.

@OFFSHARING best ideas that visit my brain, during i m walking or discussing with my friends, usually, facing the screen doesnt bring me any creative thing. about laugher no! i dont laugh when i discover or solve a serious hard question, but i feel the extreme joy and delight.

@BalarkaSen here?

Yes

I want to prove group of order 858 has normal subgroup of order 11 or 13

so here is my idea

sorry, too large for me to factorize

858 = 2 * 11 * 13 * 3

i can only count upto 10

lol

Continue.

Let $n_{11}$ and $n_{13}$ denote number of sylow 11 and sylow 13 subgroups respectively

we have $n_{11} \equiv 1(mod 11)$ and $n_{11} | 858$

the divisors of 858 are $1,2,3,6,11,13,22,26,33,39,66,78,143,286,429,858$

$n_{11}$ divides the index of the 11-sylow in $G$, which is stronger than what you stated there.

So, if the 11 sylow is nontrivial, it divides $2 \cdot 3 \cdot 13$

why ?

My birthday is today

Sylow's 3rd theorem.

(or was it 2nd? I forget)

@OFFSHARING: it is not necessary to ping my username as part of conversations that I am not taking part in - thanks

1 sec let me go back in my book

I don't quite remember

^_^

[ ]-~ (o ' ' )

Anyonere here know of any online tools or easy mathimatical to check if you have corretly inverted a prime-dual pair in linair programming?

yeah $n_{11} | {2 * 11 * 3}$

Told ya.

Did I delete the message?

yeah

@Thijser yes, you're brain. It's always online and can perform the check

2 * 11 * 3 = 66

@DanielFischer why do you delete my messages? Looks like you have a problem withthe behaviour.

@EnjoysMath yes but I need to check if my method has been correct a few times and a tool would be nice, if there is some easy math to check that would be great as well just to check for any errors

the only possible values for which this happens is if $n_{11} = 66$ or $n_{11} = 1$

@OFFSHARING No, I did. Your word choice was again inappropriate. Not bad enough to suspend, but I'd rather not leave it around.

so, suppose $n_{11} = 66$ and let us see what happens

66 is not 1 modulo 11

Be careful.

Besides the fact you're not a mathematician, and you wanted to seem the nice guy during the elections, it looks like you wanna do now what?

i mean 78

sorry

@OFFSHARING I do believe you don't intend the messages badly, but you often use words that have a rude connotation.

Daniel Fischer is a mathematician, attach whatever adjective, he's one of the best

+1.

@OFFSHARING, how come you are always fighting with someone?

@DanielFischer You shouldn't delete my messages as long as I don't break any rules. If you are not sure about what I mean simply stay away or ask me.

@OFFSHARING Being unnice to users in the chat is breaking rules. I can understand if you didn't do it intentionally.

@BalarkaSen Where I was unnice? Show me that part!

"You mean you didn't become famous enough ..." has a horribly sarcastic tone in it.

OFFSHA, you could use a few doses of category theory

@EnjoysMath Ugh.

@BalarkaSen People kidding a lot here, there was nothing sarcastic.

Horribly sarcastic? This is absolutely ridiculous.

@BalarkaSen why Ugh? It's a beautiful area of math

@BalarkaSen You also say a lot of stuff around. I don't have time to look back but you said a lot of stuff and I don't remember you ever had messages deleted.

@OFFSHARING As I said, you might not have intended it, but it was sarcastic.

@OFFSHARING If I have ever sounded rude to anybody, I do apologize.

Because that was not my intention.

@BalarkaSen You were rude to me tons of times, but I had no problem with that, I quarreled a bit with you and that's all.

I apologize :)

dudes, go to private chat

I thought this didn't have a private chat

click on someone's name and you can should be able to invite them?

@DanielFischer I don't wanna quarrel with you, but don't afford with me this stuff.

@EnjoysMath I trust you. I am not mathematically mature enough to understand the beauty, I guess.

@DanielFischer I'm neither @BalarkaSen nor other of the users here you can afford this stuff with.

I have only ever used general nonsense in serious jobs once. That is, Yoneda lemma, in proving Hopf degree theorem.

@OFFSHARING: Stop it. Now.

@BalarkaSen so if $n_{11} \neq 1$ this means $n_{11} = 78$

@JyrkiLahtonen You don't give me command, you give command when you talk to your friends and whoever you want to.

@L33ter I hope you have done the calculations right.

I will then show that $n_{13}$ in this case must be 1

Could you explain what they are doing wrong for the rest of us to understand?

@JyrkiLahtonen First greet me if you wanna talk to me. You're not my king to give me orders.

Yep

You have mod powers but you forget that you should respect people.

well $2 * 13 * 3 = 78$

and so, the $n_{11} \cong 1(mod 11)$ is $n_{11} = 78$ or $n_{11} = 1$

everybody is ignoring me today

OK, @L33ter.

Now onto $n_{13}$.

ok now I will look onto $n_{13}$

jesus christ, way too much noise

@OFFSHARING, what about making a post on meta then?

→ 23 messages moved to Trashcan

hey @BalarkaSen

so here is how I did it

I have had all my messages censored for asking a question of the mods

the only possible values for $n_{13}$ is 1 or 66

@Brennan.Tobias If you want to ask that question, the place to do it is meta.

if $n_{13} = 66$ this means there is atleast 792 elements of order 13 and since we assumed that $n_{11} = 78$, so we have atleast 780 elements of order 11

so $|G| \geq 780 + 792 = 1575$ which is a contradiction

what do you think ?

Deleted messages, unstarred messages, banning all remind me of the bad events humanity had to combat during the time.

@OFFSHARING, I would say this is bullying

Hope not to be banned one day for not having blue eyes ...

(and with this I closed my conversion)

I came here 5 minutes ago, and I only see a couple of you yelling, while a few others actually chat about a problem. Why would I think the yellers have been wronged, when they started foulmouthing others? Yelling at me is ok. I hardly ever chat, and I am just learning what a mod can do here. Don't want to use these tools, but if you make me...

Hi @JyrkiLahtonen, I asked you a question but it was removed

@Jyrki see Trashcan if you want the removed messages (link's up there ^).

@BalarkaSen ?

I didn't touch it. Apparently it was removed by a community moderator. Which is kinda telling :-)

@Brennan.Tobias Bans are generally not discussed in public while they're ongoing.

Thanks @ArtOfCode

np

@L33ter: Sounds like you were working on a Sylow theory problem. I lost the beginning. What was it?

I got the problem that I was working on but it was proving that group G of order 858 always have normal subgroup of order 11 or order 13

here is how I did it

@DanielFischer and it's also available for you. You first greet me before telling me anything else.

Question: Would a propositional logic system that doesn't use connectives nor uses any of the traditional inference rules be interesting to the math community, even if it was inefficient?

@JohnNash How are you doing?

How do I know that the solution to this differential equation exists? I know that $F$ has to be continuous for solutions to exist, and for that the denominator must not b equal to $0$ but there are many possibilities to check if $|x|<0.5$ and $|y|<0.25$. Do I check them all or is there a more efficient way? @CarlMummert

@JohnNash Yes, I want to change my name to Monica, it's far better to me. ;)

First of all I looked at $n_{11}$ and $n_{13}$ I saw that there possibilities is $n_{11} = 78$ and $n_{13} = 66$ or both of them are trivial. Assume that both aren't trivial so we will have at least 782 elements of order 13 and we will also have at least 780 elements of order 11, so $|G| > 780 + 792 = 1572$, which is a contradiction.

what do you think @JyrkiLahtonen

@L33ter: Sounds like your solution is right on the money!

good

I am currently working on this problem I am proving there is no group of order 1960 that is simple

@JohnNash I worked hard all day long. Almost forgot to eat anything.

$1960 = 2^3 * 5 * 7^2$ what would you suggest ? working on which sylow p subgroup first?

I made it that far myself. 78 and 66 are the only other options for $n_{11}$ and $n_{13}$. No room for all those Sylow subgroups :-)

yeah

What do you wanna know about groups of order 1960?

I want to prove that is not simple

@JohnNash not in details, but I already started to assign a folder where I put problems that I might include in a second book.

Yeah our prof said that as well

However, I do it slowly since I'm already extremely tired.

Depends. Many Americans I've met read it as Silov. I'm used to reading it Sülow, because in Swedish and Finnish 'y' is read as 'ü'. Not 100% about Norwegian I'm afraid.

@Paradox101: is the function that defines the ODE not continuous at the point $(-2,1)$? If that function is continuous, the ODE has at least a local solution beginning at x = -2?

@JohnNash Not sure if they allow me to do 3 uni years in one year ... and then I have to move on to the city.

well, here what I want to show one of the $n_p = 1$ and that will show that any group of order 1960 can't be simple

@JohnNash I might like to do some researching, 3 years is a lot.

@JohnNash: I don't speak French, but I'm pretty sure the 's' is silent.

@CarlMummert the ODE is continuous at $(-2,1)$. But shouldn't it be continuous within the rectangle centered at the initial condition with $|x|<0.5$ and $|y|<0.25$

Because we're supposed to check for existence of solutions within the rectangle @CarlMummert

I guess I could look at all the sylow p subgroup and assume that it is not simple and get contradiction somehow

by counting the least elements of it

@Paradox101: yes, I see what you mean. The set of points where the function is not continuous is the graph of $1/(1-x^2y^2)$. Because that graph is closed, if the function is continuous at a point, it will be continuous on some open neighborhood of the point. Basically, just draw the graph of $1/(1-x^2y^2)$ and pick a rectangle around your point that does not touch that graph.

ok I will try 7 first @JyrkiLahtonen

So a 7-Sylow is normalized by an element of order 5. But neither group of order 49 has an automorphism of order five. Therefore a 5-Sylow will COMMUTE with a 7-Sylow. That sounds like a promising start, but I don't see a route to the destination yet.

Oke I will start with you 1 second I will comment in few second after few calculations @JyrkiLahtonen

@John: true, but more interestingly the function will still be bounded below by a strictly increasing function on an interval around the point where the original function has a positive derivative, so that the increasing function is equal to the original function at the point where the derivative was taken

@JohnNash: That was news to me. I only recall having seen an example of a function that is strictly increasing, differentiable almost everywhere, but the derivative was zero (when it existed).

@CarlMummert I get the area between the blue rectangle. So then that means that the solutions exist and given that a similar condition for continuity for the partial derivative with respect to $y$ exists then the solution will also be unique right?

@JohnNash: Thanks. I did see that button right away. Decided not to use it. For exactly the reason you mentioned. I'm a bit ashamed for not really being familiar with how the chat works. So I try not to use my mod powers too much for now.

@Paradox101 Yes, that is exactly the idea. The theorem that gives uniqueness and existence is Picard's theorem. In this case, the function that defined the differential equation is a nice function, so there should be no issues with derivatives or smoothness as long as you stay away from points where the denominator is zero

Besides, I'm enjoying the last two fingerwidths of a bottle of Quarter Cask Laphroaig. One more reason not to overuse diamond powers.

@CarlMummert ok thanks a lot for your help :)

hey @JyrkiLahtonen I assume G isn't simple so $n_{p} \neq 1$ for any p

so after some computations I got that $n_5 = 56$ i.e $n_5$ is at least that

and $n_2$ is atleast 5 also $n_7 = 8$

@L33ter: Good! I think that what I said before about 5-Sylow and 7-Sylow commuting gets the job done. It means that the normalizer of a 5-Sylow has size at least $5\cdot 7^2$. So assuming that $n_7=8$ we get that $n_5$ is a factor of eight. Meaning that always either $n_7=1$ or $n_5=1$

@L33ter: Have you seen the technique of using the automorphism groups of Sylow subgroups? Here turning a normalizer into a centralizer.

oh cool let me see your way

no I haven't

I did but that was like 3 month ago

at start of semester i forgot that

@JyrkiLahtonen is it possible to calculate the order of $Syl_2(G)$ in this case

Ok. So here a 7-Sylow is either $C_{49}$ or $C_7\times C_7$. The automorphism groups are $Aut(C_{49})\cong C_{42}$ and $Aut(C_7\times C_7)\cong GL_2(F_7)$.

using the logic I just deduced

The order of $GL_2(F_7)$ is $48\cdot 42$. The key point is that in both cases $Aut(P_7)$ has no elements of order five. So if an element of order five normalizes a $P_7$, then it has to centralize it.

@Jyrki: I think my Christmas self-present will be to get a new bottle of Laphroaig 10 for my office when I get back.

oh I see

that is cool

@MikeMiller: Sláinte!

@L33ter: Implying that if $n_7=8$, then some 5-Sylow (hence all) is centralized by a 7-Sylow. Hence $C_G(P_5)$ has order at least 245. Therefore $N_G(P_5)$ has order at least 245. Therefore $n_5\le 8$.

Anyway, the punchline to that was "Enjoy!"

Oops. I was supposed to do some upvoting before the UTC day ends. Gone for a while.

1 second @JyrkiLahtonen wrapping my head around what you wrote

how do we know sylow 7 is either $C_{49}$ or $C_7 \times C_7$? @JyrkiLahtonen

pfffff, my latex file got blocked for the first time

(after trying to finish something for some time this is horrible indeed, unexpected, undesired event)

@L33ter: A group of order $p^2$ is always abelian. Those are the only alternatives.

@JyrkiLahtonen could we discuss your solution ?

I don't understand why 5-Sylow and 7-Sylow commuting ?

Make it quick. It's 1:45 am in these parts. I still have a sip remaining.

I forgot many parts related to sylow theory after I did field theory and module theory haha

so I am sorry if my questions are stupid.

Not stupid at all. And I'm "worried" that there may be a simpler way that I missed :-)

first thing I don't understand why is 7-Sylow commute with 5-Sylow

Ok. Assume that $n_7=8$. Therefore $[G:N_G(P_7)]=8$.

oh oki

alright I agree with this

So $N_G(P_7)$ is of order 245

alright

So $N_G(P_7)$ has an element of order five.

yes

If $x$ is such an element, then conjugation by $x$ is an automorphism $f:P_7\to P_7$,

alright

Because $x^5=1$ we also get $f^5=id_{P_7}$.

alright

If $P_7$ is cyclic of order 49, then $Aut(P_7)$ is cyclic of order 42. So in this case $f$ must be the identity mapping. Because a group of order 42 cannot have elements of order five.

The only element $f$ of $Aut(C_{49})$ that satisfies the equation $f^5$ is the identity mapping.

yes

And if $P_7$ is a direct product $C_7\times C_7$, then similarly $Aut(P_7)=GL_2(F_7)$ is of order $48\cdot 42$. And the same argument works.

oh I see

ok ok cool thanks a lot

Great!