You are just rephrasing.

@JonasTeuwen Common. I can say two spaces are equal, homeomorphic or homotopy equivalent. They all very different notions.

@JonasTeuwen $(0,1)\notin \langle (1,0) \rangle$

$\Bbb{S}^n$ is not homeomorphic to $\Bbb{R}^n$

@BenjaLim But you can't seem to say then what it is.

@JonasTeuwen two sets are equal if they contain the same elements.

I know that an apple is not a pear.

Then you did not tell me what an apple is yet.

@JonasTeuwen basic set theory jonas :)

Only 'not pear'.

I very much know that right.

I am just protesting against your seemingly 'rigorous' definitions.

Yes, right.

I asked my question then

@JonasTeuwen i'm not really sure what point you're trying to make, to be honest

@AlexanderGruber If I take a group of one element, and name the element 'apple' and otherwise 'pear'. Is it still the same group?

The point I am trying to make is that '=' is not purely equal as sets, but has many meanings, it is just an equivalence relation which captures the 'equality' your are interested in.

@JonasTeuwen you mean if you take the sets $\{apple\}$ and the set $\{pear\}$ with the trivial binary operation?

@JonasTeuwen they are isomorphic, not equal

If I now name you John, then you are not the same person anymore?

@JonasTeuwen no then you can say Alexander Gruber = John because you made the name

That's an assignment? I just define it.

@JonasTeuwen in group theory $=$ does denote set equality. (and i think in general algebra but i can't speak for sure.)

you define John := Alexander Gruber?

Those things can still be the same set theoretically.

I just have another bounded variable.

Equality depends on what you look at.

@JonasTeuwen i mean, i understand that you are saying writing the $=$ sign implies some equivalence relation. i'm just trying to tell you what that equivalence relation is in algebra, and why we insist on $\cong$.

I have seen other algebraists do it differently.

So perhaps, you should say who the 'we' is you are speaking of.

@JonasTeuwen they must have been speaking informally... i would never see any algebraist say $\langle (0,1)\rangle = \langle (1,0) \rangle$ as in the earlier example.

That's because it's a lame example.

I hate the word 'informally' there.

It is not informal.

@JonasTeuwen just incorrect?

It assumes that the reader knows what it means.

It's not incorrect either.

In analysis you can write $A = B$ for operators.

@JonasTeuwen what are operators in analysis?

Which can mean a couple of things: assumes same values on intersection of domains or are the same with same domains.

Functions between two spaces.

Why did I waste my time trying to help this guy math.stackexchange.com/questions/329911/…

@caveman I don't know. Have a beer.

@JonasTeuwen those aren't just functions?

So, now tell me. How would you write the distinction without too many words?

(forgive me, i'm not being deliberately ignorant. i am genuinely bad at analysis.)

They are, between topological vector spaces or whatever.

ah right okay

Alright, $f = g$. Here $f$ and $g$ are functions.

It can mean the have the same domain and coincide on that set.

It can also mean they coincide on a subset - perhaps the intersection.

Should be clear from the rest of the text.

oh right i see what you're saying

It is not incorrect.

Introducing extra symbols for things people are quite able to follow is useless.

annoying

the reason that is different between our disciplines is that algebra is mostly about the $=$ sign, figuring out equalities and isomorphisms and whatnot

it's about the nature of the objects we're working with, whereas in analysis there are a lot of more specific properties you study

So is analysis, but reducing the question to one of $\leq$ or $\geq$ is better.

That's not true.

Perhaps in the analysis you know.

i think in the specific example you gave me it is.

So?

Analysis is much bigger than that specific example.

sure, but you provided that as an example of the $=$ sign usage in general analysis

I have seen in Fourier analysis on quantum groups commutator diagrams to prove the existence of the Fourier transform, but no way of computing it for specific functions.

Yes - ...

As you declared you did not know much analysis.

Also, I have seen yesterday a very nice example of using only inequalities to bootstrap another inequality to obtain $=$.

yeah, i don't mean that all of analysis is inequalities.

but you must admit there is a difference in the nature of these branches of math - it makes sense that we use $=$ differently. i don't think either use is necessarily better, just adapted to fit the situation

Have you guys heard of "shut up and calculate"?

have i ever, @cyrillicalphabet.

Of course there is, and hanging so much on a rather vague concept suggests they are less talented.

@JonasTeuwen algebraists?

I've heard some algebraist sigh when there was one of these wise noses that wanted the $\sim$ on top of the $=$ during his presentation.

In analysis they also use $\cong$ or $=$ for isomorphism.

@JonasTeuwen that's because a presentation is not meant to be formally correct.

ok people seem to like my question guess its ok

It has absolutely nothing to do with formal or not formal.

It's just notation.

It's like saying an apple or more formal than a pear.

That most people do it that way is a reason to do it yes, but that has nothing to do with formality but with consistency.

we have agreed on a formal way of doing things to standardize the language

Yes, but not about the bloody letters.

There is a difference between a language and the letters you use to write it down with.

Or the handwriting.

Or the pen.

of course: but the language of mathematics crucially depends on the letters.

This is more like the TeX font you use to typeset it if you wish to pursue such analogy.

naw, i don't think so. you're making it sound aesthetic

the difference is a change in meaning

Also saying that the difference between analysis and algebra is like $=$ vs $\leq$...

that's just friendly horseplay, @JonasTeuwen. i'm sorry if you took offense to it.

Alright, if it really is that way, then I guess there is truth in the statement that analysts are more talented than algebraists.

I don't see it as horseplay, more like trolling.

hi @amWhy

haha, okay. well take it however you like then.

if you're going to get offended when i'm not trying to be offensive there's really nothing i can do about that, man.

@κρανίοπεριπολία Hello!

wtf?

How can it not be? I put some effort in explaining and you seem to turn my words around.

There are plenty other things I could have done.

If I immediately would have assumed 'idiot' like the other guy there above, I would not have done that.

But anyways - no offense taken.

Now, I am thirsty.

beer?

No.

Bruichladdich.

@JonasTeuwen i see what you're trying to say. you don't seem to see what i am saying, but that is fine, i'm not really all that worried about it.

@caveman i read your question. it is a good question.

thanks

Please vote for caveman's question:

@caveman i'll develop this into an answer if i can, but i believe the reason has to do with something called Chermak-Delgado measure. if we can prove that the Sylow $2$-subgroup $P$ must be centralized by (or equal to) more than $|G|/|P|$ elements in a group whose order is of that form, i'll be able to tell you why.

oh sweet!

@caveman we may be able to use my friend george glauberman's theorem to prove it, if i can't figure it out i will ask him.

it looks like I wil have to work very hard to understand the answer :D

@caveman nonabelian simple groups are horrific beasts.

but they so interesting : )

they are.