$$\prod_{k=0}^{\infty} \left(1+\frac{1}{2^{2^k}-1}\right)=\frac{1}{2}+\sum_{k=0}^{\infty} \frac{1}{\displaystyle \prod_{j=0}^{k-1} (2^{2^j}-1)}$$

@robjohn What do you make of these answers, pure nonsense just for kicks, right?

@pjs36 They do it so that people will talk about them, especially in math chatrooms

Interesting, @PaulPlummer, you seem to have a lot of insight into the minds of these delinquents.

:)

@pjs36 he was referring to the fact that you were talking about them in a math chatroom...

And I was trying to keep up the banter!

@anon He was referring to me having such knowledge sort of implies I am a delinquent

@AlexClark @BalarkaSen It is up. I spent less time writing it up carefully, let me know if you find errors, math salad, or word salad. Also I was maybe a little to informal for some peoples taste for parts of my proof.

@MikeMiller Above

It would be awesome if a spider fell right onto your head as soon as you read it...

Did you make your site with Jekyll or something similar, @Paul?

I made it with Jekyl @SohamChowdhury

@SohamChowdhury mathsalad.com/about

Mine is too, but I haven't posted much there yet.

Sweet domain name. Do you have github hosting it?

Yeah.

Very nice theme too

I have not decided whether or not I want comments...

It was originally meant to be an Aluffi blog, like Alex's DF blog.

:P

Did you finish the scheme, I have messed a bit with haskell (not much)

No, I have a lot on my plate right now :(

I got a bit farther, but not much.

I found a DF blog with like 11 chapters of solutions... that's a crazy thing to do!

Some people type faster than they write.

(or they can't read their own handwriting)

so that's a really smart thing to do in that case.

Well you can (I guess everyone can) let me know what you think of posts, since I don't have comments up right now.

@Paul: I'll take a look this weekend.

Did a spider land on top of you? @MikeMiller

Nope

In Algebra: Chapter 0 Aluffi says "If a function is injective but not surjective, then it will not have a right-inverse, and it will necessarily have more than one left-inverse." Is this true? I seem to be able to come up with a function which is injective but not surjective, and which has exactly one left-inverse: $f : \{ 1 \} \to \{ 2, 3 \}$, $x \mapsto 2$.

Hello

@flakmonkey Look at the errata for the book. It mentions this: "and if the source has >2 elements".

@flakmonkey you're right, the domain must have more than one element for there to be multiple left-inverses

sorry, I forgot to check the errata first! thanks guys

are you working through it too, @flak?

@anon: when does $a \equiv b \mod n \implies a^t \equiv b^t \mod n$?

always

but the converse isn't true, right?

right

except if?

a, n are coprime?

well, if (ab,n)=1 and (t,phi(n))=1 then the converse implication holds

okay

@SohamChowdhury yes but I'm also trying to work through some of Pugh's Real Mathematical Analysis this summer so I'm moving pretty slow

think of the integers mod n as a finite ring whose elements are all zero divisors and roots of unity

it's Munkres (with a little Simmons to start with) for me :)

with mth roots of unity, a^t=b^t doesn't generally imply a=b unless (t,m)=1

unless I finish groups and get to rings/modules.

which is the next chapter.

@PaulPlummer: I might try to write the sequel to my original post this weekend, if I still remember the math involved. :P

Can't you just let the pants guide you, @MikeMiller?

That is good, I really liked your first post, it gave me a feeling of what is going on, way out there in $>3$ dimensional land. (I know it is just a taste) @MikeMiller

@Paul: But nothing like =4 land... that's all >4 :)

I'm realizing what a shame it is that I haven't formally learned about modules very much.

I'm getting the impression those guys have all the fun.

Is it anything special that all elements are self-inverse in $(\Bbb{Z}/12\Bbb{Z})^{\times}$?

(I think they are)

It's somewhat special, groups of this form can be characterized nicely (although you'll have to wait for it). For example, you can deduce that it's Abelian.

At the very least, it's a very rare property for a group to have

@SohamChowdhury in fact all elements of $(\Bbb Z/d\Bbb Z)^\times$ are self inverse if and only if $d\mid 24$. This fact seems to be related to lattices, string theory and monstrous moonshine, an area in between modular forms and the representation theory of sporadic simple groups. See here

My goodness!

Can I prove that or does it take high-powered tools I don't have?

oh no you can prove it

I recall that the 24th power of the Dedekind $\eta$ function is special or something.

yep, like I said modular forms

Wow.

Fascinating.

there are an impressive number of downvotes on the front page at this moment

Is it okay to still be in shock that such a simple fact (which I can understand) is so connected to so many other things?

:)

Let me try to prove that.

@anon, can the composition of two set-functions $\phi: G\rightarrow H, \varphi: H\rightarrow K$ be a homomorphism if neither of $\phi, \varphi$ are?

yes

Hey, @AlexC!

What've you been doing?

@SohamChowdhury let $G\xrightarrow{\alpha}H$ and $H\xrightarrow{\beta}K$ be group homomorphisms and let $\gamma$ be a permutation of $H$ which is not a group automorphism. Then let $\phi:G\xrightarrow{\alpha}H\xrightarrow{\gamma}H$ and $\varphi:H\xrightarrow{\gamma^{-1}}H\xrightarrow{\beta}K$

I did another set of exercises, @AlexC. :)

@anon Let me parse that.

@anon Oh, that is so cool!

I was thinking of something along those lines.

iow $\phi$ is a hom plus a jumble and $\varphi$ unjumbles that and is another hom.

@SohamChowdhury right

From what I've seen from @AlexC, I have a sort of feeling that all this ring stuff is waaay hard. :)

@AlexClark it'd be simpler to show $\Bbb Z[x]/(p)$ is an integral domain

since it's just $\Bbb F_p[x]$

if you really want to do nuts and bolts, you can write $a(x)b(x)=pc(x)$, compare leading coefficients and then induct on degree

but that's a lot more complicated than what I suggested

@AlexClark we "salvage" arguments that originally had errors, we don't salvage arguments we don't know how to continue

I just mean that the word "salvage" does not work that way, not that you should give up on arguments you don't know how to finish.

@anon, @AlexC, do most algebra textbooks not have these? They are non-categorical, so I was wondering.

it depends how introductory the book is.

But this is just a way to visualize the fact that the composition of two homs is another?

It doesn't have anything to do with category theory, so I thought other books might have these too.

I like "geometric" sorts of books with nice diagrams and everything, really. Not good at forming nice mental pictures without a little help. This book is wonderful. (@AlexC . . . post-finals and -project, you know what to do)

it looks like it's writing down the conditions for $\varphi$ and $\psi$ to be group homomorphisms in the category of sets using the multiplciation maps $m$, and then putting the two diagrams together (not sure why)

@anon He says this: "the outer rectangle has to commute if the inner two do"

And then proves that $(\psi \circ \varphi)(a \cdot b) = (\psi \circ \varphi)(a) \cdot (\psi \circ \varphi)(b)$.

so yes, it's also showing the composition of group homomorphisms is another group homomorphism

personally I don't see the need for that diagram. diagrams should be used when they make life easier, not more complicated.

like writing down what it means to be a group object seems pointless if you're just defining a group, but then if you want to define coalgebras and corepresentations and so forth it makes sense to write down the idea of an algebra/representation with diagrams and then reverse the arrows.

I just wish he made it look more like the wedge of cheese that it is :(

@MikeMiller I guess what I mean by that it gives a feel for how wild 4 is, by sort of explaining why >4 is easier.

@anon group objects are the 10th section of the chapter. this is 3.

@AlexClark hello

he went off because he can't stand you.

like yesterday.

your 'ol' reminded me of this

You will be buried alive.

@anon nevertheless, it feels like it helps, at least at this moment. maybe I'll understand why you say so later.

not as obvious

@SohamChowdhury helps with what exactly?

what's your idea?

@anon with understanding how the homs "send" things from one group to another. anyway, let's not argue about this, you're way, way more experienced and obviously have very valid reasons for what you're saying.

@AlexClark so you're using the fact Z[x] is a ufd?

that's fine

Fair enough @PaulPlummer

Tried a freska ice cream float, very tasty

You're a bad human being. Google the Kolkata temperature to find out why.

Yikes, you should get one of those floats, it will be refreshing

I need one, true.

Know anything about coproducts in ${\sf Grp}$?

I know a little

I know nothing, enlighten me.

They are basically free groups, but not :D

That's very helpful. /s

free products, not free groups

They are free products

that's what they're called, the book says, but it stops there.

where can i learn a bit more?

That is why I said 'but not"

you can learn what coproducts and free products are on wikipedia

okay

What do you want to know about them? @SohamChowdhury

Given groups G and H, the coproduct C is a group equipped with maps G->C and H->C such that any two outgoing maps G,H->Y factor through some map C->Y. (Look at the diagram to make sense of this,)

anything beyond just the name. i know the universal they satisfy, but I want to know a concrete construction.

okay.

The concrete construction is very similiar to free groups,

so do you get free groups?

yes, somewhat.

by picking the identity map G->G and trivial map H->G (so our Y is just G itself here), there must me a map C->G such that G->C->G is the identity. thus, G->C is an embedding.

embedding?

injective map, 1-1, putting a copy of one thing inside the other

How do you feel about presentations? @SohamChowdhury

thus, the coproduct C has a copy of G in it and a copy of H in it.

so you can multiply elements of G and H together in C, so to speak

the question is, are there any nontrivial relations satisfied by the elements of G and H? (i.e. relations that one cannot simply deduce by simplifying products of things in G or simplifying products of things in H)

@anon oh.

and this is where the explicit construction comes into play

@PaulPlummer barely know them, haven't studied them yet. i'm only comfortable with them in the cases of $S_n$ and $F_n$.

@anon just like the disjoint union

but i know that doesn't work as a coproduct here

Basically in $G*H$, the free product, you are allowed to multiply the elements of $G$ and $H$ together to form arbitrary products of their elements (declaring that the identities of both are the same, but otherwise all of the other elements are distinct), and the only thing you're allowed to do to simplify such products is simplify things like $g_1g_2$ when $g_1,g_2\in G$ and $h_1h_2$ when $h_1,h_2\in H$

Oh, that is indeed very similar to free groups!

It's like a "formal product" of a sort then?

yes

hmm.

indeed $F_n*F_m\cong F_{n+m}$

@anon how does the explicit construction of the coprod help in finding "nontrivial relations" between the elements of G and H?

@anon I would imagine that is fairly easy to show, but I'm not sure.

@SohamChowdhury I didn't say it helps us finding nontrivial relations between elements of G and H (there are no nontrivial relations), I said that's the stage in our discussion where we're going to use this explicit construction (the free product) and show that it is the coproduct

I don't really get what you mean by there being nontrivial relations between the elements of two different groups.

(assuming they existed)

@SohamChowdhury when you finally get the hang of things it should be intuitively obvious

You might want to look at some of this here, chapter 2 talks a little bit about some constructions. There might be some examples you are unfamiliar with, but I think you might like it, plus it talks about a few other things

@anon it is already, but I'm not sure about proving that.

@PaulPlummer right, doing that (edit: that is some beautiful LaTeX!)

@SohamChowdhury say that $g_1g_2=g_3$ holds true in $G$. Then (viewing $G$ as sitting inside the coproduct) it also holds true in the coproduct. Same idea for elements of $H$. I call these relations trivial, because we know that they must be true right off that bat. The question is, can something like $g_1h_1\cdots g_kh_k$ (with nontrivial $g_1,\cdots,g_k\in G$, $h_1,\cdots,h_k\in H$) turn out to be the identity? The answer is, no, there are no such nontrivial relationships.

This site has been deleted for low quality content.

@anon i'd prefer the comments section were. one can dream.

No, math.se.

If things could be deleted for their low quality content, the internet would not exist

How can that even be possible? There is no operation defined $G \times H \rightarrow \text{something}$.

So say $g_1h_1$ is a reduced word in the free product.

How could it ever be equal to the identity?

Basically, in free products things are not the identity, unless they are obviously the identity

(I am playing fast a loose with that, since you could do free products of groups where you can not solve the word problem)

Yes, exactly my point.

@PaulPlummer "undecidable in general" I believe?

@SohamChowdhury by definition, it can't. (unless $g_1$ and $h_1$ are both the identity of course). But in general if two groups $G$ and $H$ are subgroups of a bigger group, there can be nontrivial relations between them, so it's not automatic that $G$ and $H$ sit so "freely" and "independently" inside the coproduct; one must prove this fact.

so far I've only talked about free products and coproducts as different things, and haven't offered any proof that they are one and the same (in Grp anyway)

Even then, the operation inside the free product is not the same are the one in the "bigger group", so how can you reduce $g_1h_1$ further in that group?

@SohamChowdhury huh?

the operation in (/on/inside/what preposition works here?) the free product is concatenation followed by reduction, right?

Here's what we have so far: (1) in the free product, G*H, the elements of G and H have no nontrivial relations. (2) in general if two groups G and H are subgroups of a bigger group K, it's possible for elements of G and H to have nontrivial relations, (3) the coproduct $G\sqcup H$ is a group that contains G and H as subgroups.

I mean, say G and H are subgroups of $K$.

@PaulPlummer Stack Exchange is noah's ark :)

Let $g_1 \bullet h_1 = k$, where $\bullet$ is the operation in K.

But I can't reduce the word $g_1h_1$ to $k$ in $G * H$ anyway.

[My (2) above should say that if groups G and H are subgroups of K and their intersection is the identity.]

@SohamChowdhury What is your question? Which of my three facts, (1), (2), (3) above would you like clarification on?

1 and 2.

I'll explain my question.

What is the operation in the free product?

you compose two elements of G using G's operation, you compose two elements of H using H's operation, and otherwise you concatenate (imposing that G and H have distinct elements except for their identities)

He has not defined it, but it is just concatenation, and then you can combine terms of their own group.

Yes.

yes

So, suppose G and H are subgroups of K.

That's how I defined it.

And there's a nontrivial relation $g\bullet_\text Kh = k$ in K.

@SohamChowdhury sure

And, wlog, $k \in G$.

no

not possible?

oh, yes.

i see.

then h is also in G.

right?

right

but then the word $k$ is not in $G * H$, right?

$k$ is not a string, it is an element of $K$

you obtained $k$ by multiplying an element of $G$ by an element of $H$, but a string is a formal product

yes, so it is not in $G*H$ because it is an element of neither.

nor is it a formal product of things in them, right

then it is (seemingly trivially) impossible to show that $g$ and $h$ have a nontrivial relation by considering the free product.

because you can't reduce the word $gh$ any further.

you mean $g$ and $h$ can't have a nontrivial relation in $K$? or in $G*H$?

i don't think they can in the latter

If you can't reduce a word any further then it is not trivial... (provided the reduced word is not the identity)

I already defined $G*H$ so that the strings don't reduce except in the trivial ways. That's how I defined it!

wait, wait.

@anon can $g$ and $h$ have some nontrivial relation in $G*H$?

no. I defined $G*H$ so that there are no strings that reduce nontrivially

so why did you say this?

I would imagine that fact is very obvious.

Imagine what's very obvious?

that is what confused me.

when you say "that fact," what fact are you referring to that you think is obvious?

the one you refer to at the end.

it seems like it follows from the definition of $G*H$.

$G*H$ is the free product. The coproduct is $G\sqcup H$.

oh.

I haven't offered any proof they're the same yet.

damn.

that's why.

i'm sorry for wasting so much of your time :(

should've read carefully

thanks, @anon.

mmhmm

can I show that the free product satisfies a universal property?

exactly!

like we did for the direct product and the product (in the book)?

right.

@anon: will the universal property be similar to the one the disjoint union satisfies in ${\sf Set}$?

yes

same one

:-)

@DavidWheeler is on main but not on chat :-(

Its been a long time since I met him

@Rememberme It sort of seems like as soon as I started to get active on chat, he basically left and when Jasper left @SohamChowdhury came (and brought grammar correcting tendencies with him)

spelling more often

:D

and I stop if someone doesn't like it

:)

yo @AlexC

how are the assignments going?

@Rememberme Is it hot where you are at?

Yes, very.

Season three of Star Trek has some blow your brains out bad episodes... Luckily the first episode of the season is so bad that you don't feel anything afterwords.

>2K people have died across India this year because of heatstrokes and shit.

:(

@SohamChowdhury Seems like you guys can not catch a break, earthquakes, hot weather. But doesn't normally get hot, during some parts of the year, was it it causing so many problems now?

that is so cool.

I didn't know that last one, about the hyperbolic plane.

@PaulPlummer yeah, but some years it's worse. rain is of course a huge factor and there's not been much this year (till now at least).