That axiom makes a lot of sense, at least in words. Those logical symbols are way too pretentious, although they might be useful in conserving space. Continuous prose is the way forward.

You'll get used to them sooner or later.

@Khallil There is a reason for using logical symbols. Ultimately, logic has to be formalised in the study of mathematical logic.

@Khallil However, other than the purpose of studying mathematical logic, math texts should be written in ordinary prose.

Agreed. It's much more comprehensible to beginners that way. Although in a sense, we're all beginners, haha!

I should be paid 1 million USD for my consultation above. =)

There's a sysadmin in the room!! Run for your lives!

I've got a quick question on subsets. Would it be correct to say that $\{ 1 \}$, $\{ 2 \}$, $\{ 1,2 \}$, $\{ \varnothing \}$ and $\varnothing$ are all subsets of $\{ 1, 2, \varnothing \}$?

Who cares? At most get a lifelong suspension, lol.

@Khallil Yes.

What's a sysadmin? A system admin? Even if that's right, I still don't get the joke. Oh! HAHA!

@Khallil That is correct.

Wow, I'm slow!

It's not even a joke, lol.

@Khallil Did you try out the problem I gave you?

@BalarkaSen Tried and failed as I had to go and do some shopping, after which I totally forgot about it!

@robjohn Yeap.

@BalarkaSen Can you think of any other ring like structures which defining galois groups would make sense

@Alizter yes, that's what me and @blue were discussing lately. it turns out that we can define galois groups for pretty much anything, even for categories.

@Chris'ssis adding the $[n-(n-1)]$ would make the proof clearer IMO

for what it's worth, i was fiddling with galois groups defined for groups.

so i have no reason to think that it won't work for rings.

@robjohn Yeah. The point is that this time I didn't give many explanations in that proof. I usually provide with more details.

@Khallil Try it now then.

@Alizter Aut_H(G) are automorphisms of G fixing H pointwise. Characteristic subgroups you can google to see what they are.

@BalarkaSen Do you know of any papers or formal research addressing the subject?

@Alizter No.

None.

I only know that galois categories were formalized by Grothendiek who used it to pose the so-called geometric galois theory. look for SGA1.

Talk to me about the group one

@Alizter what do you want to know? it's still under construction.

Strange that you kids are talking about super advanced stuff.

I guess you have started with field extension like constructions?

@Alizter no, i haven't used field extensions. just the usual group inclusions.

So take a (finite) group $G$ and a normal subgroup $H$ of $G$. And then a galois like extension $G|H$?

normal subgroups isn't the correct way to analogize galois extensions. the mimicry is done by characteristic subgroups (google it)

take a subgroup H of G and define Aut_H(G) to be the automorphisms of G which fix H pointwise. that's what you do in fields, right?

@BalarkaSen I imagine you have a moustache. Is that right?

haha, no, i don't.

Hahahaha

@BalarkaSen Any subgroup?

@Alizter for now.

@JasperLoy What's the point in worrying about what others are doing? Focus on yourself!

@Khallil Well said. I wasn't worrying though.

Mhm. A better word would be concern.

That sounds a bit rude. Hmm, it's hard to find the right word.

@BalarkaSen Lets take cyclic groups as an example

All normal subgroups are characteristic

sure, take it.

I don't like to use the word rude, because it is not clear what it means

So do the automorphisms form a group all the time or do we need to be more specific on the type of subgroup?

Is $\varnothing$ a subset of itself?

@Khallil Yes.

I thought so.

@Alizter Prove that Aut_H(G) is a group. That's an exercise for you.

@Khallil Every set is a subset of itself.

A subset of another set is one whose elements are totally contained within another, right?

Yes.

One would think that Balarka is a professor giving out exercises.

Then no elements would be contained by every other set. Which is the same as saying that $\varnothing$ is a subset of every set (and without avoiding redundancy, itself).

@JasperLoy If I am offending people by giving out exercises, I do apologize. I was under the impression that people enjoy doing problems.

@BalarkaSen No, no. That was just a casual remark.

If only we can stop hair growing on the wrong parts of the body, lol.

I hate to have to shave.

@Alizter As Aut_H(G) is a subset of Aut(G), elements satisfy the associative property and are left unchanged while composing with the identity. You are only left with the existence of inverses.

@BalarkaSen Not at all, it's good fun trying your (well at least they're yours by extension) problems out. ^_^

@BalarkaSen Automorphisms are bijective therefore there is an inverse. Then show that the inverse is part of the group right?

These are standard results, so one really should study X and Y first before studying Z.

yes, that inverse of automorphisms that fix H pointwise also fix H pointwise.

@JasperLoy X and Y?

@BalarkaSen You know what I mean.

Yeah. I guess I do.

@BalarkaSen All inverses of auts of G are also auts of G

It's similar here.

@Alizter How do you prove that Galois groups are groups?

@Alizter That's not enough.

If they leave h fixed then they are their own inverses

wat

wait

wat

first tell me how you prove galois groups are groups.

Yes thats how

where's the proof?

@BalarkaSen By noting the 'group' in Galois groups.

(^_^)"

@Khallil That's like saying an open interval is open, lol.

@BalarkaSen $\varphi(x)=x,\;x=\varphi^{-1}(x)$

Therefore it has inverses

@JasperLoy Hahahahaha! It is though, isn't it? ;-)

what the heck is $\varphi$?

@Khallil Yes, but it needs to be proven, because open has a certain meaning.

be explicit.

@BalarkaSen an aut that fixes h

@JasperLoy Open. I'm pulling your leg! I understand what you mean. ^_^

@Khallil What if I have no legs? =)

@JasperLoy I'm sorry if I've offended you, or anybody else reading this.

@Alizter Right. An aut $\varphi$ which satisfies $\varphi(x) = x$ restricted to $H$ must have an inverse preserving $H$ as $\varphi^{-1}(x) = x$.

@Khallil What? No need to say sorry.

@BalarkaSen Thus proven

@Alizter It always helps to be explicit in your proofs.

@BalarkaSen I am typing in chat

Thus proven. Now, is there anything stopping you to do that in the case of Aut_H(G)?

@BalarkaSen My proofs are explicit

Ahem. I am typing in chat too.

Anyway, let's get to the math.

@Khallil Is Ted still angry with you?

@BalarkaSen No

@JasperLoy Not at all. I think my hiatus from here has managed to smooth things over.

Exactly. So Aut_H(G) is a group too.

$\{ X \subseteq \mathbb{N} : | X | \leqslant 1 \} = \{ \{1\}, \{2\}, \{3\}, \{4\}, \dots \}$, right?

@Alizter Anything else you want to ask about?

@Khallil Oh, a closed interval is also closed, and that also needs proof!

@BalarkaSen Now what can we say about Aut_H(G) with respect to G?

@JasperLoy Noted. ^_^ I think I've seen the proofs of that nature somewhere.

Lets compute for $C_4$

I am not interested in computing.

Have you read my claim?

@Khallil You missed out the empty set.

@BalarkaSen link

@JasperLoy Oh, right!

That is precisely a mimicry of the fundamental theorem.

@JasperLoy What I wrote above would be right if the cardinality of $X$ was equal to $1$, right?

@Khallil Yes.

@BalarkaSen Any proof?

@Alizter I got one.

You interested?

@BalarkaSen Good :)

@JasperLoy If you're asked for a set of all subsets of $A$ such that the set of all subsets must have members with cardinality greater than that of $A$, the set is then empty, right? For example, $\{ X : X \subseteq \{ 3, 2, a \} \text{ and } |X| = 4 \} = \varnothing$.

@BalarkaSen Yes. Ping me when you have it all typed up

I need to eat.

I will look at it after :)

OK.

I have it posted somewhere, so let me find it.

@MikeMiller

The Galois correspondence is between the set of subgroups of G and subgroups of Aut(G), if I am right, right? The partial orders are the inclusion homomorphisms and the order reversing maps are $\text{Aut}_{(-)}(G)$ and $\text{Fix}(-)$

which, composed together in any order, returns the identity map in the corresponding set.

The essence of the material world is language, the essence of language is mathematics, and the essence of mathematics is nonsense.

@BalarkaSen Could you drop me a cryptic hint to do with determining the truth of the statement $\mathbb{R}^2 \subseteq \mathbb{R}^3$? I know that it's true just by looking at 3 dimensional space totally containing 2 dimensional space, but I can't seem to justify it in terms of ordered pairs. Is it as simple as deducing that $(x_1, x_2, x_3) = (x_1, x_2)$ where $x_3 = 0$?

@Khallil Identify $(x, y)$ with $(x, y, 0)$ in $\Bbb R^3$

Then the inclusion follows.

@BalarkaSen Oh nice! It did seem obvious, but I couldn't determine whether that statement needed a proof of it's own.

Notice that $(x, y) \neq (x, y, 0)$. You are just identifying one with another.

@Khallil $\Bbb R^2$ is isomorphic to a subset of $\Bbb R^3$

Right.

Technically speaking R^2 is never a subset of R^3, but there is a copy of R^2 sitting inside R^3.

@Alizter That just seems like jibber jabber to someone (me) who hasn't learnt the definition of isomorphic.

@Balarka Well, recall that one of your subgroups has to be characteristic. In that sense you might well call it a sort of "group extension" G over H, obviously not in the standard sense.

But yes, what you sai sounds right, though admittedly I&' not thinking very hard

@MikeMiller sai? I&'?

said, I'm

The problem with thinking G over H as an 'extension' (in the nonclassical sense) is that we can't explicit construct the extension, as we can do with fields. there is no sense of basis while working with groups.

plus, characteristicness of inclusions doesn't identify with extensions but with galois extensions.

Sure, it was just phraseology. Of course there's no mathematics behind what I like to call it. :P

I'm not used to Big Oh notation. Is the following true $$O(\frac{1}{(n+1)^2}) - O(\frac{1}{n^2}) = O(\frac{1}{n^2})$$? :)

@BalarkaSen How were you technically speaking? The book I'm working with says $\mathbb{R}^2 \subseteq \mathbb{R}^3$ is false, but doesn't explain why.

@Khallil Ah, then it clears matters up.

Consider an elt of R^2. Say (1, 2). Is this in R^3?

It's not, hence there is no inclusion.

@Chris'ssis does the same method work for $$\sum_{n=1}^\infty\psi'(n)^2$$?

@BalarkaSen Yes! That's because $(1,2) \neq (1,2,0)$ right?

Exactly.

Got it!

A more appropriate reason would be that all the ordered sets in R^3 consists of 3 elements while (1, 2) is an order pair.

In general, you can embed R^2 in R^3 in a lot of ways, by identifying (x, y) with (x, y, 0); by identifying (x, y) with (x, y, 1), etc.

@Khallil Here's a reveal for the problem I gave you : $x \in (A \cup B)' \Rightarrow x \notin A \cup B \Rightarrow (x \notin A) \join (x \notin B) \Rightarrow (x \in A') \join (x \in B') \Rightarrow x \in A' \cap B'$. This shows $(A \cup B)' \subseteq A' \cap B'$. But then $x \in A' \cap B' \Rightarrow (x \in A') \join (x \in B') \Rightarrow (x \notin A) \join (x \notin B) \Rightarrow x \notin A \cup B \Rightarrow x \notin (A \cup B)'$, which implies $A' \cap B' \subseteq (A \cup B)'$.

Hence you are done.

Duh, I forgot the latex symbol for and.

@BalarkaSen $x \in (A \cup B)' \Rightarrow x \notin A \cup B \Rightarrow (x \notin A) \land (x \notin B) \Rightarrow (x \in A') \land (x \in B') \Rightarrow x \in A' \cap B'$. This shows $(A \cup B)' \subseteq A' \cap B'$. But then $x \in A' \cap B' \Rightarrow (x \in A') \land (x \in B') \Rightarrow (x \notin A) \land (x \notin B) \Rightarrow x \notin A \cup B \Rightarrow x \notin (A \cup B)'$, which implies $A' \cap B' \subseteq (A \cup B)'$

ah, ok.

Sorry, the \join errors made it hard to read.

I'm still trying to understand it.

(It was \land, as I'm sure you already know!)

no, I forgot it.

I have a poor memory.

Relative to Von Neumann, we all have a poor memory!

^_^

@robjohn Yes. Now you can see why I can do it wpp. :D (back from jogging)

hi, please help me, what does mean a name of category followed by an arrow pointing down like this: $\mathrm{Top}\downarrow$?

@BalarkaSen I've got a problem with the last part before the final implication. $x \not\in A \cup B \implies x \not \in (A \cup B)'$

@Khallil what's the problem?

I meant $x \notin A \cup B \Rightarrow x \in (A \cup B)'$

@BalarkaSen Yea, that's what I thought!

@Khallil For what it's worth, it's called DeMorgan's law in set theory.

@BalarkaSen The name sounds familiar. I might've looked at it somewhere before.

Oh, and the two things we've proved end up showing that if the two sets are subsets of each other, they're equal.

Yes, and I believe that you can prove it easily?

Haven't we spoken of this before?

Hmm?

I remember talking about something similar to proving that if two sets are subsets of each other, then they're equal.

I don't =P

Bad memory, eh? Haha! ^_^

Chernoff and his silly bounds.. I shouldn't have went with proving every single statement in the paper

@BalarkaSen Did you find it?

@Alizter Find what?

Your proof

What proof?

Oh, that one.

Yeah, let me find it. Apparently I forgot.

Well, on the other hand it wouldn't take much time to write it up either.

@Alizter The exact sequence was 1 --> Aut_H(G) --> Aut_N(G) --> Aut_N(H), correct?

something like that

OK, recall that N is a subgroup of H and H is a subgroup of G. Is it obvious to you that 1 --> Aut_H(G) --> Aut_N(G)?

Note that automorphisms of G which fixes H automatically fixes N, as N is a subgroup of H.

yes

So it's straightforward that Aut_H(G) is contained in Aut_N(G), i.e., 1 --> Aut_H(G) --> Aut_N(H) is exact.

The exactness at the third morphism is what we want now.

@BalarkaSen I am not too familiar with exact sequences but I am trying to follow.

@Alizter you know what they are, right?

Something to do with chains of homomorphs

and the image of the hom before is the kernal of the next

Right. So 1 --> Aut_H(G) --> Aut_N(G) means that the kernel of the morphism Aut_H(G) --> Aut_N(G) is trivial, i.e., it's injective - hence Aut_H(G) is contained in Aut_N(G). Did you follow upto that?

hmmm, another awesome proof just came to mind ... to another problem ...

@robjohn btw, did you try the cubic version?

@BalarkaSen yes

@Alizter Now start with the map Aut(G) --> Aut(H), defined by restricting all automorphisms to H. This is well defined, as H is a characteristic subgroup of G.

Yes

@Alizter Now restrict the domain of the map Aut(G) --> Aut(H) to Aut_N(G). It follows that the restricted map is Aut_N(G) --> Aut_N(H).

any answers?

@Alizter Can you determine the kernel of this map?

@Scorpion I can help you get there, but I won't give you the answer straight away.

@Khallil that's cool, thanks

@Scorpion If 10 people carry 2 boxes each, how many boxes are being carried?

20

ah I think I see where this is going

@Scorpion Yep, that's the same as 10x2.

@BalarkaSen Well it is trivial right?

@Alizter The fun thing is that I was told the same by a person some while back. Why should it be trivial?

@Khallil let me try this out

@Khallil so is it just 5x * 8y = 161 ?

@Scorpion Why did you multiply the two together? That equality is for the total number of boxes being carried.

If 10 people carry 12 boxes each and 5 people carry 10 boxes each, the total number of boxes, $b$, is equal to $(10\times 12) + (5\times 10)$.

meant to write +

@BalarkaSen Well if auts of G fix N then the Auts of H also fix n. The zero of G is inherited to N therefore if it is being fixed it has to be the zero element therefore trivial...?

wat

@Scorpion Oh, ok. Yea, that's right! Now can you deduce anything about how many people there are altogether?

you are not making sense

sorry I have math anxiety @Khallil

hmm

not sure how to work that out

@BalarkaSen The kernel of $f$ is the set of elements such that $f(x)=0$

Yes, ok.

So?

Therefore x has to be zero if f fixes x

@Alizter What is the 0 here?

the identity hom

of what?

N

Be explicit, otherwise you'll make silly mistakes.

@Alizter Nopes.

Let me write this

The map was Aut_N(G) --> Aut_N(H)

@Scorpion Read the question again. There are 25 people altogether. $x$ people are blue and $y$ people are red. Try and form the equality like you did for the number of boxes being carried.

hmmm

@BalarkaSen I am thinking wrong wait

were $N\subset H \subset G$ subject to any extra conditions?

yes. those are all characteristic inclusions.

N char H char G

@Khallil to find how many people carried 8 boxes I would put y=1/8(161-5x)?

i.e. making y subject

@Scorpion Woah, steady on. You're only making things more tedious. We've got our first equality, $5x + 8y = 161$. We're also told that there are 25 people altogether, $x$ of which are blue, and $y$ of which are red. This means that the total number of people present is equal to?

25

@Scorpion We already know that it's equal to 25 which also happens to be equal to?

@BalarkaSen If it is a characteristic subgroup should that not mean that Aut_N(H)=Aut(H)?

@Alizter Eh, why is that?

Well if every aut of H fixes every element of N by definition of char subgroup then should that not mean every aut that fixes n is every aut of H?

but you asked people present, which is 25? no?

or am I understanding something else

@Alizter you're mixing up invariance and pointwise invariance.

Ah @BalarkaSen Kinda like a neighbour hood

ahahahhhahah

I am silly

Hi @Nimza :-)

@Scorpion There are $x$ blue people and $y$ red people. How many people are there all together?

xy

if there are 2 blue people and 3 red people, are there a total of 2 * 3 = 6 people?

Of course. Number Six wants to join this interesting mix of colours.

you're ignoring the multitudes of people who aren't bizarre colors

@Alizter Complete the exercise.

$G\backslash H \cup \{0\}$

you think that's the kernel?

Depends I am having trouble thinking about the homomorphism\

Is it surjective?

No.

@BalarkaSen I can't see it being nontrivial

@Alizter The map is Aut_N(G) --> Aut_N(H). The kernel, by definition, is the group of elements of Aut_N(G) which, restricted to Aut_N(H), goes to the identity map -- i.e., fixes elements of H pointwise. Thus it's Aut_H(G).

@BalarkaSen My head hurts

I'm a bit spooked by the passage in Book of Proof by Hammond concerning $\mathcal{P} ( \mathbb{R}^2 )$, the power set of $\mathbb{R}^2$.

I get it but this is all very nonobvious to me

it depends on how much you've worked with morphisms. if not much, it'd surely hurt your brain. if you get some time later on, try drawing a diagram and understanding it diagrammatically.

@Khallil what about it?

@BalarkaSen I can only do it justice by linking it to you via image.

OK

What's spooky about it?

Oh, and try the homework. =P

That's it's that incomprehensibly big.

$\mathcal{P}(A) = \{ X : X \subseteq A \}$ where $A = \mathbb{R}^2$.

@Khallil 'course it's big. when you learn about cardinality, you'll see that it has cardinality $2^\mathfrak{c}$

It's something much larger than the cardinality of the continuum.

@BalarkaSen So $\mathfrak{c}$ is the cardinality of the continuum? Also, WOW!

Yes. It's the cardinality of the reals.

or the cardinality of P(N), to visualize it's hugeness.

And we're supposed to study this stuff? Sounds interesting!

Very interesting!

@Hamou

@Khallil Can you compute the cardinality of |P(A)|?

for some finite set A

@BalarkaSen I've seen a few proofs of it, but I can't remember them. However, I do know that $\left| \mathcal{P}(A) \right| = 2^{|A|}$ for a finite set $A$.

Yes. Try to prove it.

Shouldn't be too hard for you.

@BalarkaSen Ok, fine!

By induction?

That's the most natural form of proof for me right now, as I've practiced it the most.

@Khallil Hmm, I am not sure about induction. Do you know a little bit of counting? i.e., binomial coefficients and whatnots?

@BalarkaSen Not much other than the fact that $\displaystyle \binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ for computational purposes only.

Ah, then you don't know that $\binom{n}{r}$ represents the number of subsets of cardinality $r$ of a set of cardinality $n$?

Nope!

Ah, OK. Nevermind, move on. I'll have to think about a completely elementary proof, if there is any.

anything wrong with this? drive.google.com/file/d/0By4cQB8gmcRjTWFLcERoU1lVQ2c/…

@Chris'ssis I have not, but I figure it adds one more layer to the process.

@robjohn That version is much harder. It cannot be easily tamed.

@Khallil Aha! I have it!

@Chris'ssis Is it? I will have to try the same methods on that when I get a chance.