Mathematics - 2016-10-13 (page 4 of 5)

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4:00 PM

@MaryStar yeah isnt that a solution?

For all $z \neq 0$?

I got the maps $\psi_i D_{2n} \rightarrow D_{2n}$ given by $b \mapsto ba^{i}$ and $a \mapsto a$

we can explicitly define it with its principle value to make the function work

this map

@shaihorowitz For $y=0$ we get $x=\pm 1$.

4:01 PM

@Adeek that says how Z_n acts on D_2n. now how does Aut(Z_n)?

$\theta_k : D_{2n} \rightarrow D_{2n}$ given by $a \mapsto a^k$ and $b \mapsto b$ where gcd(k,n) = 1

Yes. Can we also check if it has solutions for $x,y\neq 0$ ? @shaihorowitz

right

but that how can I get the explicit map @arctictern ?

@MaryStar all solutions are of the form y = sqrt(1-x^2)/sqrt(2) so we need to design x to make the right side rational

4:03 PM

@MikeMiller @MikeMiller I've a guess that, there is no continuous map $\mathbb RP^3 \to S^1\times \mathbb RP^2$ which induced the inclusion $Z_2\to Z\times Z_2$

@Adeek if you know how $\Bbb Z_n$ and ${\rm Aut}(\Bbb Z_n)$ go into ${\rm Aut}(D_{2n})$, then you already know how $\Bbb Z_n\rtimes{\rm Aut}(\Bbb Z_n)$ goes into ${\rm Aut}(D_{2n})$. you have to check that it's actually a group homomorphism of course.

@Anubhav That's not correct. There's a map $\Bbb{RP}^3 \to \Bbb{RP}^2$ that does that; it's the quotient of the Hopf fibration by the antipodal map (upstairs and downstairs).

@MikeMiller Ohh yes...

hm I don't understand @arctictern

@Adeek consider $\Bbb Z_2\times \Bbb Z_2\to \Bbb C^\times$ where $(1,0)$ maps to $\alpha$ and $(0,1)$ maps to $\beta$. can you tell me where $(1,1)$ is mapped to?

it's like that, except not commutative now

4:07 PM

it will be mapped to $\alpha*\beta$

@MaryStar check this out math.stackexchange.com/questions/738446/…

just a sec washroom

i have to go see you guys later

See you!! @shaihorowitz

Bob Dylan - Death Of Emmett Till

4:11 PM

@arctictern so I know how $\Z_n$ and I know how $Aut(Z_n)$ goes into $Aut(D_2n)$

is it just the composition ?

@Danu YCor is a more frequent MathOverflow user than MSE user. I think he can encourage or discourage crossposting (with attribution) as he pleases.

On the other hand, I'll have an answer soon.

@MikeMiller Crossposts are discouraged, SE-wide.

The principle is that it's not right to ask for two groups of people to spend their time on your question at the same time, when one may well be enough.

But yet, there have been many crossposts throught the SE community.

Yeah, which I regard as a bad thing.

Not a huge deal, and of course anyone is allowed to disagree.

Shrug. I think it's fine to ask an MO question once asked in MSE and if the people there cannot answer.

4:18 PM

Yeah, of course

What is the difference between MO and MSE?

Once you've established that, though, you should flag to migrate the question.

The fact that you wait until you're sure-ish that the MSE community will not help (in the short term) already means you've done what I try to promote

NB: I am stupid. It is actually easier than expected...

user image

Hooray for the left zero semigroup eating away all potentially problematic terms

(This example is still rather trivial through)

This might be a stupid question

@Danu MO is mostly allowed to do what it likes. It's on the SE network to use the software, and not so much to abide by community rules.

4:26 PM

@MikeMiller But we're on MSE.

But what is the geometrical interpretation of a complex contour integral?

In either case, I don't care that much. I'm just defending my actions as normal and consistent. As I said, feel free to disagree.

I'm not going to waste my time on this.

exactly

In comparison, the geometrical interpretation of a Riemann integral is the area under the graph

4:26 PM

Hi @Ted.

Is there a roughly equivalent way of seeing things in the case of complex contour integration?

Hi, @Balarka. Have you slept?

hey @Ted!

No, @Lozansky, it's more like the interpretation of a line integral as giving work of a force field.

Took a smallish nap, yes, but should still be able to fix the sleep cycle.

4:27 PM

hi @Danu

@Balarka: It would be easier for you just to adopt the Krebs cycle.

I forgot what that is. I dropped biology in 10th.

Photosynthesis ... I never took biology until late in college, but we didn't do Krebs cycles.

Ah, right.

We had to memorize the whole ATP-generation process... ugh. Lots of names.

@TedShifrin Okay yeah I guess that might be a satisfying answer

@MaryStar x=199/201

4:29 PM

That's the best we can do, @Lozansky :)

@Balarka: You an expert on principal and asymptotic directions yet? :)

hey @TedShifrin

hi Karim

I computed the Gaussian curvature of loads of surfaces, while half-asleep (that's the braggable part!).

I was solving the following every linearly indepedent set cant be extended to a basis of F

where F is free abelian group

I understand principal directions now.

4:31 PM

It kinda weird how certain things in linear algebra

doesn't generalize to free abelian groups

Yes, modules are more subtle than vector spaces.

@Balarka, since you expect this of me, I'll ask: As a matter of principle? :)

yes

most things in linear algebra generalize to modules

(free finite type ones)

@TedShifrin I wanted to ask you yesterday actually. I m supposed to take like ring and modules next semester and

this is one thing that doesn't

but most do

4:33 PM

Agreed, @JuanFran.

representation theory for my third class I can choose between functional analysis, algebraic topology, differential geometry/mechanics, functional analysis

what do you think I should take ?

and complex analysis

@TedShifrin The earlier version of me appreciates that pun. The most recent version wants to say that was a flop.

@Balarka: The most recent is so high-falluting and discerning?

@Anubhav, @BalarkaSen: Answer is posted.

All of 'em, Karim ...

4:34 PM

@MikeMiller Cool, I'll read.

i agree, all of them are fundamental and useful

What magnum opus this time, @MikeM?

I can only take 3 classes as full time student @TedShifrin

Complex analysis is absolutely fundamental, next comes algebraic topology. I don't know what differential geometry/mechanics means.

it studies mechanics using differential geometry

4:35 PM

@TedShifrin It has grown a better taste for puns.

@TedShifrin Here. The OP asked if there were maps between the fundamental groups of 3-manifolds that were not induced by continuous maps.

surely you have some preference adeek? are you more interested in Analysis or topology/geometry

Oh, is every homomorphism representable for the functor or something, @MikeM?

@Balarka: I remain to be convinced.

@TedShifrin ualberta.ca/science/-/media/science/departments/math/…

@TedShifrin What do you mean by representable? (It's true if we're looking at maps to K(G, 1).)

4:37 PM

@TedShifrin There's no functors here. This is 3-manifold topology.

Unless you're super-interested in physics, Karim, that's low priority to me. Learning actual differential geometry (not just the definition of a manifold) would be high priority.

I was thinking the functor from spaces to groups, that's all.

I am an iota insulted that Moishe thinks there's nobody on MSE that can answer the question.

Asking if every group homomorphism is in the "image" of the functor. I wasn't being technical with the use of the word.

I don't know who Moishe is.

Me neither.

I might make statements like that occasionally, but you never know if Robert Bryant might suddenly appear like a deus ex machina ...

4:39 PM

your use of "iota" is interesting

It's a common use, @Balarka. iota for small bit ...

Analysts would use epsilon. :D

Oh, that's for the old Balarka.

I don't think I have heard it often.

@Ted: I try my best to solve sufficiently interesting MSE problems.

I've had to work very hard to solve a few here ... Just saying ...

Me too.

4:43 PM

@BalarkaSen Well, iota is the smallest lower-case Greek letter. Tho I don't think anybody's ever used it as a variable... so it's confined to prose.

2+2=5

@J.M.: I use $\iota$ frequently for involutions of various sorts.

@TedShifrin Usually, really fun problems are hard. :)

@PhysicsGuy I assume you have evidence?

@J.M. Fair enough.

4:44 PM

@TedShifrin The lower-case version?

Absitively.

@TedShifrin No, i just wanted a trolly way to say hello

Hello everyone

I don't usually say hello to trolls.

Unless they're under nice bridges.

I'm not a troll.

Maybe I should rewrite the answer to give a more general collection. Maybe that's not necessary.

4:46 PM

@TedShifrin You'll want to take a billy goat with you in such a case. :)

Billy goats are usually reserved for mountain escapades.

I guess I pretty much write the general conclusion at the bottom.

We'll see if the seven people who liked the question come back.

@PhysicsGuy I suppose you'll claim that 2+2+2+... = -1 next?

Your mother's so fat, she can't be embedded in R^n

@MikeMiller I should plan a couple days on Christmas in which I'll just sit down and read the (more interesting) topology answers you have posted in MSE :)

4:47 PM

Just a joke

I don't know that it'll be a fun Christmas for you.

I suggest watching Mr. Robot instead. Pretty good show.

@MikeM: I guess that answer is a lot easier if we work in homology rather than $\pi_1$.

Every map on H_1 is probably induced.

Oh, no.

My proof shows that's not true.

@Physics Guy every compact manifold can be embedded in R^

so no

every homology class is induced comes from a submanifold

@MikeMiller I don't watch shows, but Huy recommended agents of shield once.

4:51 PM

You certainly can't get every map on $H_2$.

@JuanFran The compactness assumption is not necessary.

@Ted: My answer gives many examples where all maps must be degree zero.

@TedShifrin Right. Every map $S^2 \to T^2$ has degree zero.

With a good cohomology structure on the domain, you get many examples.

Right, @MikeM, but we can concoct that just for surfaces.

Balarka just did.

4:53 PM

In a test on topology: "Are the topological spaces X and Y homeomorphic ?" "X is, but Y isn't".

Surfaces are easy. 3-manifolds are harder.

@JuanFran That was a "Yo mama" joke.

I think we should consign @PhysicsGuy to the "questionable humor" room.

I had to do some work because you cannot conclude that just from the ring structure.

I understand, @MikeM.

4:53 PM

Every map $S^3 \to T^3$ is nullhomotopic, doesn't that mean it's zero on $H_2$?

Um, most maps are 0 on the 0 group.

@TedShifrin "Questionable humour room" ?

Ops.

Go back to diff geo, @Balarka :P

Perhaps the question should be not “Do good math jokes exist”, but “are they unique”?

4:55 PM

@TedShifrin I have proved that $Aut(D_{2n|})$ isomorphic to $Z_n \rtimes Aut(Z_n)$ but I would like to compute the explicit map that does this ?

Meh, I am dumb.

is it just the composition

69

A: Mathematical "urban legends"

This is a story that I heard from one of the postdocs from my university, which in turn heard it from one of the professor at the university (I didn't bother to verify with him as the source seems relatively reliable). The said professor was a postdoc in some university in the USA a few decades ...

I mean the way I proved that $Aut(D_{2n})$ isomorphic to

Karim: I'm not thinking about algebra today.

4:56 PM

okay

@Ted: I gave him my KW question yesterday. But he doesn't have the obstructions to say anything about it yet.

What do you get when you cross a chicken with an elephant?
The trivial elephant bundle on a chicken.

That one's good.

@MikeM: Indeed not, and he won't learn about abstract Riemannian metrics in those notes. Everything (except for $\Bbb H^2$) is sitting in $\Bbb R^3$.

@Ted @Mike: Yeah, if I have an example for H_2 I would automatically have an example for H_1 by Poincare duality so that's not any easier...

You're on 3-manifolds, then, @Balarka?

5:00 PM

I am slow today. I'll get back to diffgeo.

@TedShifrin Right, 3.

No you wouldn't. Poincare duality is not natural unless the map is degree 1.

@MikeMiller Ah, excellent point.

The obvious example is $S^2 \times S^1$, projection to a factor, then inclusion.

Is there a theorem that says that an nth order ODE needs $n$ solutions to form a fundamental set of solutions?

@Lozansky: Linear ODE?

5:02 PM

Yeah

Sure.

Ah cruds.

The solution space is an $n$-dimensional vector space, @Lozansky.

You can find proofs lots of places (including a couple of my books). :)

@Ted: I'm sure he can figure the way things work abstractly.

The stories are similar.

@MikeM, but for KW you have to think about flows on the space of Riemannian metrics.

5:03 PM

@TedShifrin Thanks! :)

Shh.

For embedded metrics we have obstructions beyond GB. They're called Mainardi-Codazzi equations :P

I know. I'm more excited by the abstract question, though. I prefer intrinsic geometry.

Beautiful application of Frobenius to prove they (and Gauss) are necessary and sufficient.

I guess I actually have spent most of my life doing extrinsic geometry in one way or another, so we're again on opposite sides of the fence.

OK, really back to reading diffgeom now. I wish I was less dumb today.

5:08 PM

@Ted Occasionally I'm fond of using extrinsic geometry to better undersfand the total space. Thinking calibrated submanifolds.

Calibrations are totally extrinsic, yes.

Andrew Thompson

I need help picking one from the following list: beer, rest and math.

Yes, but take eg an almost complex symplectic manifold - a (generic) count of curves gives GW = invariant of symplectic manifold.

Similarly with more exotic structures.

Hell, or instantons.

@MikeM: I just went and checked. Every single one of my published papers is about extrinsic geometry.

Is "rest and math" one element in that list?

5:10 PM

I was wondering the same thing, @Lozansky :)

@AndrewThompson What time is it?

Andrew Thompson

7PM.

The importance, yet again, of the Oxford (or serial) comma.

Beer and math.

That was not one of the two choices!

Andrew Thompson

5:10 PM

Not applicable. And no, three distinct items.

The extra comma would make three distinct terms :)

You didn't say Mike had to choose one unique element.

Oh, you did say one. My apologies.

Andrew Thompson

Yes. I agree the "and" could cause confusion, though.

With the proper extra comma, there's no ambiguity. :P

Do it anyway.

Andrew Thompson

Yesh, been three years since I've had to explain things to robots (i.e. programming.)

Well, the beer comes with social interaction, combining that with math is difficult, even at the bar of the math department.

5:13 PM

The math department has its own bar? Cooool....

Andrew Thompson

It used to be very cool, and then we moved. Now it's a classroom with a fridge and a bartable.

Oh, not cooool ...

lol

Andrew Thompson

Nevertheless, cheap beer by Norwegian standards.

@AndrewT I disagree. I end up talking about math at all of these things.

Get your beer.

5:15 PM

I actually learned a lot of math as a grad student when a bunch of us went for beer and pretzels Friday afternoons after geometry seminar.

Andrew Thompson

If there's only people interested in mathematics, sure. Sad thing is; not all mathstudents at the undergraduate level are.

Or maybe I just thought I did.

The Kirby students drew lots of pictures.

Andrew Thompson

Butyeah, I think beer wins. Bye!

@TedShifrin Why do we call asymptotic curves asymptotic?

@Ted When was this?

5:18 PM

ok, maybe for a not so profound reason

I guess you had the tiniest overlap with Danny R, huh?

One like. I shouldn't have spent those hours on it. Let the little shit crosspost.

5:36 PM

@MikeMiller I'll read your answer, currently I'm reading a survey on 3-manifold by Hatcher... I'll come back to your post after finishing it...

@Anubhav You're fine :) I'm complaining at the wind.

Good evening

6:19 PM

Hi @Alessandro

@Alessandro How are you?

@MikeMiller Have you considered posting on MO? I feel like the level of your contributions is very high---which may actually cause them to not get much attention on this site.

@BalarkaSen See exercises 21, 22.

Greetings @Alessandro

Hello @TedShifrin !!! I have a question.

Let $L=\{ (x,y): x^2+y^2<5\}$.
Suppose that we have a function $u(x,y)$ that is harmonic in $L$ and $\Omega \subset L$ an arbitrary space such that $(0,0)$ belongs to that space.
Suppose that $u$ is equal to $1$ at the boundary of $\Omega$.
How can we compute the integral $\int_{x^2+y^2 \leq 4} u(x,y) dxdy$ ?
I thought that we could use the second version of the mean value theorem.
The theorem is the following:

If $u$ is harmonic in $\Omega$ and $\overline{x} \in \overline{\Omega} \setminus \partial{\Omega}$ then $u(\overline{x})=\frac{n}{ w_n r^n} \int_{|\overline{x}-\overline{\xi}| \leq …

Why don't you just post on the main site?

It seems like you might as well post such a worked out text as a question, no?

@Evinda: I don't see how to do anything unless you know $\Omega$ is the ball of radius $2$. But in general you should switch the double integral to polar coordinates and use the mean value property.

Also, what you're doing things related to complex analysis, you should never use bars unless you mean complex conjugates.

6:35 PM

@Ted How does undergraduate studies in the USA work? Was reading up on that and it seems that they hardly know any math when they finish

@Krijn: Depends very much on the university and the particular undergraduates.

But, yes, we award a very great many weak degrees. On the other hand, most of the students who took my Honors multivariable math class as first-year students ended up doing very solid degrees (many including graduate courses).

I got interested by a question on Quora about math on MIT

Which said he finished his math degree as an undergraduate with eight courses in maths.

MIT is one of the more structured, demanding programs. But plenty of places offer 15 different degree programs, some more CS, some more bio, some more stat, etc.

I did an undergrad degree at MIT, and took something like 10 graduate courses.

But, yes, eight courses above calculus and differential equations is probably all that's required.

Remember that in the US only a tiny percentage (overall, less than 5%) of math majors go on to graduate school.

So they miss out on all the beauty of maths

Not all.

But relatively few people want to make math research their lives.

And there's nothing wrong with that.

6:39 PM

So those that go on to graduate school are already heading towards academics then?

We should not design degree programs, even at MIT, just for those wanting to go to a top Ph.D. program. That does everyone a disservice.

Or is graduate school a valid option if you know that you want to go into business

Not necessarily going into academics, but probably the majority think they are.

Ahhh, never knew that

A few people will do a masters or a Ph.D. in math and then work in finance or something. But that's not that common. Most people will go into the "real world" right after their undergraduate.

6:41 PM

Completely the opposite from the Netherlands then

Some people who do Ph.D.s in applied areas want jobs working in industry, for sure. But a lot more of them go do Ph.D.s in CS these days, I think.

This is way too big a country, Krijn :)

There are thousands of math degrees every year.

Is that why Trump is dividing it?

LOL @Lozansky

In the Netherlands, @Ted, it's sort of the norm to continue with at least a master's.

Sorry bad joke, I know nothing of politics

6:42 PM

A PhD is a very different story in NL---this is different from the situation in the US.

He's just enabling all the haters. Most of them were there, just hiding more quietly.

Can we assume that $\Omega$ is the ball of radius $2$?
We have that $\int_{x^2+y^2 \leq 4} u(x,y) dx dy=\int_{r^2 \leq 4} u(r \cos{\theta}, r \sin{\theta}) r dr d{\theta}$. Right?
But how can we use now the mean value property?

By the way @Ted

@Danu, we also have plenty of students who go from math into other areas (law, medicine, biostatistics, ...)

I am writing the notes for my seminar presentation (to be held in a few weeks)

6:43 PM

I have no idea what you can assume, @Evinda. Ask your professor.

With regard to your question, do the $\theta$ integration first.

I'm currently wondering what the best way is to rush through the set-up of fundamental classes and PD in ~10 minutes.

Can't you assume everyone knows that, @Danu? Spend 5 minutes at most establishing notation, but don't try to teach all the foundations they should all know.

Just remind them of the salient facts, establishing notation. Don't derive, don't explain, unless someone asks.

@TedShifrin Well, I did get the explicit assignment to at least "recall" pages 270--end of appendix A

it is so cold here

I hate it when buses are late

I'll have about 1.5 pages of notes on it

6:46 PM

If you do all 1.5 pages of notes, that'll take half the hour.

I've got 90 minutes, btw

You think my tempo will be that slow?

Oh, ok. My advice is to sketch out the whole talk and then figure out what to skip or do very lightly. Remember to do plenty of examples, examples, examples.

Just rushing through stuff at breakneck speed doesn't help. The people who know it are bored. The people who don't know it get lost immediately, anyhow.

Right now, I have: 1. recall $H_n(M,M-x)=\Bbb Z$, 2. define local orientation, define global orientation. 3. quote theorem that gives us fundamental class 4. remark on different coefficients and manifolds with boundary

Ultimately, the question is what your audience actually knows. I have no clue.

Then PD: 1. define compactly supported cohomology 2. cap product definition 3. quote PD

@TedShifrin It's a bit tricky... Most of the audience will have taken the topology that I have

So that'd stuff they all already know

6:49 PM

@TedShifrin The limit of integration does not depend on $\theta$.
Oh, I think it should be $\int_{x^2+y^2 \leq 4} u(x,y) dx dy=\int_{0 \leq r \leq 2} \int_0^{2 \pi} u(r \cos{\theta}, r \sin{\theta}) r dr d{\theta}$, right?

Is compactly supported cohomology really necessary for the rest of the talk?

but there are 3 who won't have, and I don't know how "fresh" it is in their minds

I was amazed throughout my career that many people who gave colloquia on algebraic geometry type stuff spent several minutes (re)teaching the audience what projective space was. One should assume a certain basic background, even in colloquia. Just not too much.

@BalarkaSen To define PD? :P

Work with compact manifolds?

6:50 PM

Not on compact manifolds.

Balarka and I are in accord on this one.

Sure...

but I'm supposed to set up this stuff for all the talks to come

If you do Thom isomorphism at some point, then define compactly supported whatever.

I'm off to friends

I can't guarantee the book doesn't use non-compact at some point

Have a good night/morning/afternoon

6:50 PM

Right, maybe there's something on the Euler class.

@TedShifrin Right, I'm supposed to do the set up so that those people don't have to

You can't do everything and people who don't know it can't absorb it at that pace.

I'm also doing the talk on the Euler class

I think it's better to do stuff when it's motivated to use it, not just teach a whole seminar on preliminaries.

But that'll come later

6:51 PM

There's nothing worse than sitting through a lecture that says: You'll see why this is important next month.

Hmm... Okay @Ted. The main part of my talk is section 4, which is axioms + consequences of SW classes

@TedShifrin Yeah, I got annoyed by that on the grad student seminars when I was in alggeom mode.

@Balarka: Did you see my reply re asymptotic directions?

@Danu You don't need compactly supported stuff for section 4, I can guarantee.

1 min ago, by Danu

@TedShifrin Right, I'm supposed to do the set up so that those people don't have to

6:52 PM

@TedShifrin Yup.

@Danu: I'm saying that's bad pedagogy ... and it'll all be forgotten a few weeks hence.

@BalarkaSen I've already read, typed up & understood section 4, but yeah... I can't leave others hanging :P

@TedShifrin Hmmm... But is it up to me to challenge this?

The seminar is essentially just trying to follow the book by MS

@TedShifrin Do you think most say probability theorists or combinatorists know anything about projective space?

@PVAL But they'll be lost anyway when one the topic of the talk is Weil conjectures and l-adic cohomology, so eh.

@PVAL: I think they all saw the definition in grad school, yes ... or in any number of other colloquia. I suppose we did have grad students at UGA who never took graduate topology and don't know it. But a breakneck-speed definition helps? I dunno.

6:54 PM

@BalarkaSen That probably isn't a good topic for a colloquia.

@Danu: Talk with the professor in charge. But I think it makes more sense to introduce things as they're needed, especially if they'll be forgotten if you just list them robotically in the first lecture.

I've seen plenty of colloquia of that ilk, @PVAL.

@TedShifrin I'll meet with him next week and I'll bring it up.

@TedShifrin I certainly don't know things that I think that to probability is on that level.

That's true, I suppose.

I've also seen topology colloquia where I was lost after 5 minutes, and I know some topology.

OK, I resign from this discussion. I'm all in favor of accessible colloquia. Working seminars meet a different standard.

6:56 PM

@MikeMiller taught me about dirichlet's theorem yesterday.

Happy diff geo, @Balarka. It's lunchtime for me. Bubye.

Byes!

@TedShifrin The integral becomes $\int_0^2 \int_0^{2 \pi} u(r \cos{\theta}, r \sin{\theta}) r dr d{\theta}$. But can we continue given that u is not known?

Bye @Ted

@Danu So, what's new?

7:11 PM

I tried to ask a question in the chat yesterday, it became too big. so I tried formulate a question. http://math.stackexchange.com/questions/1967248/finding-redundant-constraints
Is it suitable for MathSE? How should I tag it? Thanks!=)

7:32 PM

@Danu The questions on MO are usually out of my range. Nobody posts this sort of question on MO.

@BalarkaSen Are there any particularly good/useful exercises in chapters 0 and 1 of hatcher?

Sure.

Lots of 'em in 1. Also some in 0.

7:52 PM

@Adeek Suppose $G=HK=\{hk:h\in H,k\in K\}$ (with $H,K\le G$ subgroups), and $\phi:G\to Q$ is a group homomorphism. If you know the value of $\phi(h)$ and $\phi(k)$ for all $h\in H,k\in K$ then do you know the value of $\phi(g)$ for all $g\in G$? Yes, $\phi(hk)=\phi(h)\phi(k)$ since $\phi$ is a group homomorphism. (This makes no assumption that $HK$ is an internal direct product or that $h,k$ commute.)

oh I see

oh I see

thanks a lot @arctictern

brb class

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