Mathematics

1:03 AM

@anon given a sylow p subgroup of G of prime order not power of the prime is intersection of two sylow p subgroup then trivial ?

anon

@KarimMansour what would you guess?

btw sylow theory is irrelevant to your question

Karim Mansour

I would guess yes they should be trivial

I am trying to prove it atm

since they are of prime order so they are cyclic

by lagrange theorem

so suppose x and y are generators those two p-subgroups respectively

anon

Suppose H and K are different subgroups of G, and each of them have no nontrivial proper subgroups. What can you say about their intersection (call it M)?

Karim Mansour

their intersection must be p or 1

yeah

since the intersection is a subgroup of M and Q respectively

anon

since M is a subgroup of H, either M=1 or M=H. similarly, M=1 or M=K. so either H=K or M=1.

Karim Mansour

1:09 AM

yeah

but if the intersection is H that means both of them are equal

so yeah

same thing for M = K

so yeah good I can use that the # of non-identity elements is n_p ( p - 1)

good

DemCodeLines

2:01 AM

Anyone here good at discrete math?

skill patrol

2:16 AM

218

Q: Thinking and Explaining

How big a gap is there between how you think about mathematics and what you say to others? Do you say what you're thinking? Please give either personal examples of how your thoughts and words differ, or describe how they are connected for you. I've been fascinated by the phenomenon the que...

DemCodeLines

0

Q: Expressing propositional function without quantifiers

I have been following "Discrete Mathematics and its Applications" textbook by Rosen, 7th edition. I have come across an exercise question (1.4, #20) that I am not sure how to answer. The book gives me the question and a bunch of choices that apply, but I'm not sure how I would solve and arrive a...

discrete-mathematics

Silent

3:26 AM

@robjohn, why $S$ in bijection $f:\Bbb N\rightarrow S$ where $S\in \Bbb R$ does not always preserve well ordering?

anon

@Silent huh?

Silent

@anon, I want to understand , e.g. while N has well ordering while $\{1/n\}$ does not, why?

anon

@Silent start drawing the set {1/n} on a number line. clearly it has no minimum element.

Silent

@anon, but that i can't understand why, i mean which property a set must have to be well ordered?

anon

do you know what "well-ordered" means?

Silent

3:36 AM

yes

anon

then you know what property a set must have to be well-ordered

get a feel for how ordinals looks. all well-orders are essentially ordinal numbers.

dunno what else to tell you

Silent

ok thanks

and all ordinal numbers are well ordered?@anon

anon

look them up

Silent

ok

Shahab

I am looking for an interview of a famous mathematician (perhaps it was Atiyah) where he propogated the thesis that the reason for doing maths was two fold: getting new definitions (making theories) and proving theorems on them. He then goes on to compare the relative importance of each.

Karim Mansour

3:45 AM

if we have

nvm it is true

if we have a subgroup of normal group, then it will be normal.

no

what I am saying isn't true

1 second

Akiva Weinberger columbus

4:28 AM

@Silent A set is well-ordered if every subset of it has a minimum. So every finite set is well-ordered. $\Bbb N$ is well-ordered. $\{0,1,2,3,\dots,0',1',2',3',\dots\}$ is also well-ordered, where we define $0<1<2<3<\dots<0'<1'<2'<3'<\dots$.

@Silent $\Bbb Z$ isn't well-ordered, as it doesn't have a minimum element. $[0,\infty)$, the set of nonnegative real numbers, isn't well-ordered, as the subset $(0,1)$ has no minimum element.

The Substitute

4:45 AM

off topic. is there a way to plot multiple (overlapping) graphs in WolframAlpha?

nevermind doh

Martin Sleziak

9:38 AM

When I saw this post in the close review queue and saw that for the most part it consist of the picture I have edited it. I was doing so in good faith that I am helping a relatively new user.

Only then I noticed that the OP was not seen at the site since 2014. This was a bit annoying - I basically spent my time in wain.

I am used to seeing mostly recent question in the close votes review queue. (Although if some users vote to close old posts which are really bad in order to clean up the site, it is a commendable effort.)

Next time I should keep in mind to look on the age of the question, too.

Agawa001

10:41 AM

no interesting questions these days

this seems a good one but fits to be in different stack-section

Understand

11:44 AM

Can anyone help me understand what a prime subfield is? It says it is a subfield generated by the multiplicative identity

Why does that not just give back the field itself?

Hi @Rememberme can you answer my question above? I have to leave soon

Remember me

@anon Mind checking a proof of mine
Q). Prove that a star convex set is simply connected.
Consider points $a_0,a \in A$ where A is star convex. Now by the definition of star convex there exists a path between $a, a_0$ lying in A. That is,
$F(a,t)=(1-t)a+t(a_0)$. Consider a path $p:I \to A$ . Then the composition defined by $f:I \times I \to A$, $f(s,t)=F(p(s),t)$ such that $f(s,0)=p(s), f(s,1)=a_0$ is a path homotopy between the constant map and p. Since the straight line homotopy is the path homotopy here, $\pi_{1}(A,a_0)$ is trivial. Also A is path connected . These two show that A is sim…

Ghost

12:16 PM

hello?

Remember me

12:27 PM

Hey @Huy

Transcript for

$Mathematics$ Mathematics