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3:28 PM
@LeakyNun Hello here? :)
 
@KasmirKhaan hi
 
yeeey :D
Hello !
I done some exercices on iso and hom I want to see if i did them right =p
You got time ? :D
 
sure
 
Okay , f :G-->H is an iso, prove that the ord (f(x) ) = ord (x) for all x in G
assume that ord (x) =n
f(x^n) = f (e_G) = f(x) ^n = e_H
we showed in other ecxerccie that
f(x^n ) = f(x)^n
@LeakyNun that was all ><
 
@KasmirKhaan then it's wrong :)
 
3:34 PM
@LeakyNun for the first exercie that is
Hmm
why
we have isomorphism
f(e_G) = f(e_H)
that is allways the case , identity maps to identity
assumed order of x is n
so x^n = e_G
f(x^n) = f(x) ^n
proved this in previous exercice
since those are equal to the identity
the order of f(x) = n
all elemts have the same order once we have isomorphism
 
f(x)^n = e does not imply ord(f(x)) = n
 
oh right ><
I have to assume that ord (f(x) < n
then prove its a contradiction =p
assume that ord f(x) = m where m < n
that means that f(x)^m = e_H
f(x^m) = f(x) ^m = e_H
but since the map is isomorphic, only the identity maps to the identity
hence x^m = e but the order of x = n
contradiction
@LeakyNun better now ?
 
1 min ago, by Kasmir Khaan
but since the map is isomorphic, only the identity maps to the identity
why?
 
isomorphic is 1-1 between other things =P
 
use terminologies.
 
3:42 PM
f(x) = f(y) iff x=y
so we can't have e_H coming from other elemnt than e_G
 
alright
 
it's called "injective".
or "1-to-1"
 
if f: G-->H is an isomorphism , need to prove that G is abelian iff H is abelian
let x and y in G f(xy) = f(x) f(y) = f(yx) = f(y) f(x)
f(x)f(y) = f(y) f(x) first direction is proven
if H is abelian then f(x) f(y) = f(y) f(x)
that means that f(xy) = f(yx) by definiton
thus G is abelian
 
@KasmirKhaan elaborate on this step
 
3:46 PM
f(x)f(y) = f(xy) and f(y) f(x) = f(yx)
because G--> H is isomorphism so is H-->G
 
no, not this, earlier
 
f(x) , f(y) are elements of H
they commute => H is abelian
 
sure they are, but what does abelian mean?
 
xy=yx for all x,y in H
 
right, go on
 
3:48 PM
the follow up question is
assuming that the map was only homomorphism
do we need extra condition on the map for this to be true
my answer is no, we did not use that the map was isomorphic we only used hom
 
no, I mean go on with your proof
I still won't give you a tick
 
what do you mean a tick ?
I was done by that proof =P something missing ?
 
@KasmirKhaan ✓
 
ah ><
Okay hmm
 
2 mins ago, by Kasmir Khaan
xy=yx for all x,y in H
have you proved this?
 
3:50 PM
Yes f(x) f(y) are elements in H
call them a, b instead
so ab=ba for all a,b in H
 
@KasmirKhaan you haven't proved that.
 
hmm let me think
 
you really like to jump steps.
 
haha no but that was i see in the solution
but let me try to make it clearer
 
44 secs ago, by Kasmir Khaan
haha no but that was i see in the solution
you copied the solution?
 
3:52 PM
copied?
oh no no
 
I don't get that sentence
 
i meant thats how I see the solution for the exercice like
I dont see other steps needed
 
ok, continue thinking
 
hmm i cant think of more to say
like f :G-->H
we take an element from G and send it H
f(x) and f(y) are arbitrary elements in H
coming from arbitrary elements x and y in G
@LeakyNun do you mean we do the same steps but as in g : H--> G ?
we let a and b be elements in H
 
@KasmirKhaan no I don't mean that.
 
3:59 PM
then what is missing
 
8 mins ago, by Kasmir Khaan
so ab=ba for all a,b in H
prove this.
 
I did
f(x) f(y) are elemnts of H
arbitrary
 
@KasmirKhaan why?
 
because x and y were arbirray
 
...
start with "for all a,b in H".
 
4:00 PM
the map is from G--> H is injective and onto
 
@KasmirKhaan which one are we using?
 
Surjective
 
correct.
 
all elements are mapped
 
 
4:01 PM
-.-
:D
 
@KasmirKhaan you don't feel that mentioning surjectivity is necessary?
 
well you are right
prove that the multiplicative group R\ {0} and C -{0 }are not isomorphic
my idea was that C has 4 elements raised to power 4
equal to identity
where R has only 2 such distinct elements
 
ok
 
they cannot be the same group then
is that good?
 
yes, depending on the level of rigour needed
 
4:05 PM
Dont think in disproving something we need much
just something that is out of order =p
prove that the addtive groups R and Q are not isomorphic
 
@KasmirKhaan R\{0} and C\{0} are not isomorphic since πi is in C\{0} but not in R\{0}
 
my idea was that the order of R is bigger than ord (Q)
since what is in C ?
 
$\pi i$
 
oh
I could used also that no element in R
g^2 = -1
okay hmm
I think the best argument for addtive groups R and Q
 
do you think my argument is valid?
 
4:07 PM
is the argument about order
Yes because pi i squared is negative
no such element in R has that property
 
...
 
or wait
well yes it is valid
why not ?
 
then continue thinking
 
grrrrrrrr
no its not
because we can define a map from x--> ix
 
it shows that you don't understand what you need to disprove isomorphism.
 
4:11 PM
I told you my idea
4 elements raised to power 4 in C = e
also the order of C is bigger than R
many reasons why its not isi
 
@KasmirKhaan why does that disprove isomorphism?
@KasmirKhaan why?
 
because isomorphism groups have exactly the same properties
they are the same group in a sense
just different labels
well the order argument still is the strongest in these cases
because R is contained in C
and isomorphism imply 1-1 and onto
so the imaginary elements wont be hit
 
hmm
what if I told you R and C have the same order?
 
hmm is there a bijection between R and C?
that cant be the case :o
 
@Balarka: Of course, Casorati-Weierstrass is what I intended. It's easy.
 
4:16 PM
@TedShifrin Hello ! :D
 
@KasmirKhaan infinities are weird. Don't deal with them. For now.
@Ted hi
 
Okay =p I know that Z and N are considered to be the same
since we can find bijection between them
 
right
yet N is contained in Z
 
@Kasmir: Sure, there's a bijection between $\Bbb R$ and $\Bbb R^n$, just not a continuous one.
 
Yes but they are both infinlty countable
Oh hmm , did not know about this=p
C is like R^2
so that surprises me
 
4:17 PM
indeed you can't write out any bijection between R and R^n
 
C is of the form (x,y) if we forget about the element "i"
so anyways
 
7 mins ago, by Kasmir Khaan
4 elements raised to power 4 in C = e
you need to elaborate this
because I don't think you understand why it is valid
(and until you justify it, it is invalid)
 
well in R only 2 distinct elements raised to 4th power equal 1
in C we have i,-i ,1,-1
if they were isomorphic they would have the exact same structure
this is an examply where they fail to have that same structure
 
define structure
 
hmm well structure of the group, like if one is abelian the other is
classification is the correct word i think
 
4:21 PM
@LeakyNun Why not?
 
@AlessandroCodenotti :o
@KasmirKhaan I refuse to give you any more hint until you can start from definitions.
 
@LeakyNun both have the same cardinality, so a bijection exists
it's just not a nice one
 
that is by definiton of being isomorphic , the same group
@Semiclassical how dare you -.-
 
@KasmirKhaan where the hell do you see that vague definition?
 
4:23 PM
@Semiclassical why are you pinging me
of course I know a bijection exists
 
then why the :o ?
 
because you can't write out any bijection
 
admittedly, that string of steps is pretty painful looking
but it seems explicit enough.
 
nvm, I was wrong
right, you can write out an explicit bijection.
 
I'll pass on actually trying to understand said bijection, mind
 
4:26 PM
@Semiclassical I'll give you an easier bijection
 
okay
because isomorphic is 1-1
 
firstly, biject the real part and the imaginary part to [0,1)
then, interleave the digits in the decimal expansion
 
we have 4 elements in C that maps to 1 after raised to 4th power
 
after that, biject the [0,1) to the real line
 
and 2 element in R
 
4:27 PM
then you're done
(this needs some justification though)
 
that's between R^2 and R?
 
@Semiclassical yes
 
so this is not an injective map so its not isomorphic
@LeakyNun that is what you wanted?
 
that seems rather close to the second answer, though, which the author explicitly indicates is flawed.
so i find myself a bit paranoid.
 
4:28 PM
@KasmirKhaan that's just some elaborate nonsense to me.
 
@Semiclassical it is flawed
you find (0.4999..., 0.5) and (0.5, 0.5) mapping to different elements
 
there's a comment to that answer indicating how it can be repaired, but I have no judgment as to its validity
 
if we assumed the ker (f) = {e}
and we ttry to prove that f is injective
well let me tell u what i wrote
assume that f(x) = f(y) with x is not equal to y
f(xy) = f(x) f(y) = f(x) f(x)
 
what is f?
 
4:31 PM
f(xy) = f(x^2)
f :G--> H as before
a homomorphsim
 
what is G and what is H?
 
they are groups =p
 
are we still on the R and C problem?
 
no no different one
 
so you gave up that one?
 
4:33 PM
Need to prove that f is injective if and only if the kernel of f is the identity subgroup
I showed only 1 way stuck on the other
@LeakyNun did not give up,but i cant think of anything atm
I said what I had in mind
like isomorphism is by definiton the groups are the same
have any difference means not isomorphic
 
@KasmirKhaan where the fuck do you see that vague definition @_@
 
that is enuf for me
what do you mean vague
 
@KasmirKhaan it means you need to rely on gut feeling
turn open your book
 
i said they are not injective
 
see what it says about two groups being isomorphic.
 
4:35 PM
i know the definiton
bijective homorphsim
it fails to be injective
 
@KasmirKhaan why?
 
because of the case i found
4 elemts to 2 elents
the order wont match , we can have injective function when we have different number of elemnets
 
@KasmirKhaan who said they have to map to each other?
I really don't like to ask you questions like this.
 
0
Q: Show that continuity of probability for increasing sequence of sets implies the same for a decreasing sequence of sets

ALannisterI have the following definitions of continuity of the probability function: For an Increasing Sequence of Sets:Suppose that $B_{1} \subseteq B_{2} \subseteq B_{3} \subseteq \cdots$. Then, $P\left(\lim_{n \to \infty}B_{n} \right) = P\left( \cup_{i=1}^{\infty} B_{i}\right) = \lim_{n \to \infty}P(...

 
You're supposed to justify everything yourself.
but if I don't ask you
you won't justify anything
can't you just keep asking yourself why??
 
4:38 PM
its from the definiton
bijective homomorphsim
 
@KasmirKhaan but who said that those 4 elements need to map to those 2 elements?
 
i never said that
 
then why do you base your argument on that?
 
I said no other elements in R other than 1 and -1
have that property
Disproving something is done by a case
I gave that case
 
1. no other elements in R other than 1 and -1 have that property
2. there are elements in C other than 1 and -1 having that property
3. hence C and R are not isomorphic
don't you feel like you need some step between (1,2) and 3?
 
4:41 PM
hmm
 
you should do more exercises on basic proving.
 
those elements in C other than 1 and -1 wont be mapped so that the map is isomorphic
I dont really know what else to say about this
anyways tell me what is the missing step ?
 
@KasmirKhaan think about it until you know it.
1. not p(R)
2. p(C)
3. therefore, not q(R,C).
 
I did not do logic
 
It seems like this room is always a lot emptier than it ussed to be.
 
4:44 PM
@ALannister wrong timing
 
Also, very few people on the main page are looking at questions I post now.
Perhaps it's because they're about probability?
 
@KasmirKhaan 1. p is not true for R
2. p is true for C
3. therefore, q is not true for (R,C)
 
hey, I've tried getting an answer to a question quite a few times on various different places and haven't had one yet, can I ask / link a question here [depending on whatever someone prefers, a rewritten or a link]
 
@LeakyNun we are not expected to solve problems that way, we did not have logic yet
 
It may be a really poorly formed question, at least going by the lack of reaction elsewhere
 
4:45 PM
everyone should learn logic before doing maths ;_;
 
You can link it @Phase I do it all the time
 
0
Q: Laplace transform and meaning of $s$, and its link to Fourier Transforms.

PhaseI've recently been introduced to Laplace transforms, and my understanding so far is that it's a continuous analogue to a Summation of a power series, that maps injectively a function $f(t)$ to another function of a new variable $s$, $F(s)$. My question though is that if we take $$\int_0 ^\infty ...

 
@KasmirKhaan or else people will be hand-wavy in proofs
to other people: how do you deal with logic?
@robjohn
 
@LeakyNun yeah next course will be logic but we did not have it yet
'okay next question about the ker (f) = {e}
 
everyone should know logic.
 
4:53 PM
then the map is injective
 
@LeakyNun yes?
 
I proved only 1 way
 
@robjohn how do you learn logic?
 
@robjohn what is a group action?
I forgot who gave me the example you or someone else john
was with matrices and permutation
 
@KasmirKhaan 'tweren't me
 
4:54 PM
Okay =p
was like a is the vector (1 ,0,0) b = ( 0,1,0) and c ( 0,0,1)
permution those is like workin on S_3
and then called it a group action
 
@LeakyNun if it doesn't just come naturally, I guess one would take a logic course.
 
oh yeah it was anon
:D
 
Someone would like to see this question math.stackexchange.com/q/298360/56801 bumped. I don't know how other than by posting it here.
 
@LeakyNun Sort of depends on what you mean by "logic"; formal logic, or just common sense logic
 
6
Q: Game theory Computing pure Nash equilibrium probability

user61542We have a $2$-player game and each player has $n$ strategies. The payoffs for each player are in range $\left[0,1\right]$ and are selected at random. Show that the probability that this random game has a pure deterministic Nash equilibrium approaches $1-1/\mathrm e$ as $n$ goes to infinity. Can ...

 
4:56 PM
if P and Q are diagonizable operators, then is R = P+Q also a diagonisable operator??
 
that was not cool
 
@KasmirKhaan deleted
 
does nto change the fact that it is not cool
-.-
the action is done now , cant be undone
hows that for logic?
-.-
 
I've also noticed things being extremely tense in here and people constantly fighting with each other...
 
we are not its just for taking a break
anyways
 
4:58 PM
help please
 
many people don't have common sense logic
 
Let G and H be groups, let f: G--> H be a hom
 
Next time, demand trial by combat.
 
prove that f is injective iff the kernel of f is the idenity subgroup
I proved this ==>
and not so sure about <==
 

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