Puerøsola:
#groups-rings-fields
cloud walrusBOT
delicate chasm
Same thing different dress
We use multiplication by integers to denote repeated addition
sonic current
it's just notation
delicate chasm
And exponentiation to denote repeated multiplication
delicate chasm
If you like $x^n$ doesn't mean $n$ acting on $x$.. it just means $x$ multiplied by itself $n$ times
cloud walrusBOT
Puerøsola:
bleak abyss
@sonic current that is some impressive dedication
delicate chasm
Hi Dami
sonic current
i take it personally
bleak abyss
If I fail at most tasks I basically give up immediately
solar wyvern
@bleak abyss i am surprised yes still at it 2
delicate chasm
Persistance is good
sonic current
seriously
you'd do better if you just copied the proof and went for other exercises
delicate chasm
Please remind me never to go against you in an endurance exercise yes
I'm too lazy
sonic current
there are infinite different conditions for a subgroup
and you're getting stuck in one which is really not the hardest
delicate chasm
Wewww
bleak abyss
I mean, there are bread and butter arguments that you want to know
Either you have to figure it out yourself or you have to see someone else figure it out but internalize the fact that these are ways you should approach things
sonic current
jesus we're still at it, okay
bleak abyss
If there's an exam on the line especially, you may not want to get too caught up in rediscovering everything for yourself. It's perhaps better to figure things out on your own but make sure you have enough proof techniques when the exam hits, whatever that takes
sonic current
closure is usually done after proving the subset is not empty
@chilly ocean, focus
closure is usually done after proving the subset is not empty
so, in this case, you take any two x,y in the subset, and prove the result of their operation xy is also in H
what is the operation of the group?
a subgroup will always have the same operation as the group
can you post the whole exercise, btw?
in any case, it's usually nothing too fancy and you can just assume it's multiplicative in notation
cloud walrusBOT
creamy shits:
sonic current
exactly
cloud walrusBOT
creamy shits:
sonic current
in that case, h^k = e only guarantees k is a multiple of the order
suppose x is such that its order is two. then, x^4 = e
no
it is not
that is exactly what i'm telling you not to assume
x^4 = (x^2)^2 = e^2 = e
but the order is not 4
yes
a multiple of the order
solar wyvern
I think i might have assumed you knew what order meant
sonic current
the order is the least positive integer n such that, for a given x, x^n = e
and no, not every element in H
the order of any element in H
no
read the exercise
yeah!
which is very different from elements that divide n
we've been through that
so, point being:
we want to prove that, for any x, y in H, xy will also have the same property necessary to belonging in H
solar wyvern
(also show e has the property that order of e divides n for all n)
sonic current
it's been done
solar wyvern
sonic current
...one hour ago
sonic current
so, @chilly ocean, as any standard proof we want to use our hypothesis
which in this case is that x, y belong to H
@solar wyvern thanks
follow me
so, @yes, as any standard proof we want to use our hypothesis
which in this case is that x, y belong to H
it's the other way around though
the orders of x and y divide n
no
you know that the order of x divides n, because x is in H (your hypothesis)
no
so building on that, you want to show that xy is in H
what is H again?
prove that in an abelian group G the set of elements whose order divides a fixed integer n is a subgroup
ok
let me sketch something
that might not be irrelevant
abelian is important for stuff like
(ab)^2
which is a^2b^2 iff it is abelian
(ab)^2 = (ab)(ab)
i don't think so, maybe you'll do it now
most likely
xy=yx won't be important, you'll be using the abelian for the exponentiation
ok I think this might work
if a|c and b|c
hmm
just a sec
i think we need to raise xy to an arbitrary power again
but say, if n is 6
and o(x) = 3 and o(y) = 2
well yeah
cloud walrusBOT
creamy shits:
sonic current
that's probably how it's done
yes
I was trying to figure out what the exponent would be
see, you did it
that's a weird way to show appreciation for your own effort
but to each its own
come on
what is the problem now
a few
just do another one but quicker
you're still grasping the basics
when's the exam?
we're on the same boat
i'm studying for complex analysis from scratch too
covert vector
yoooo
sonic current
well just do a good job on the groups, ring theory will follow easily
covert vector
I studied complex analysis 1 day before my exam
covert vector
I went to a total of about 2 hours of lectures
sonic current
fuck off 😦
covert vector
I went to the first, left after an hour cuz she kept getting distracted from the smell outside or something
then the next class spent more time motivating why we might care about complex analysis, then I left in like 20 minutes
then couple months into class my friend convinced me to see if the lectures have gotten any better so I went a third time, but they were still bad
sonic current
well i'm barely starting
hopefully I'll do it in time
@chilly ocean, hit another exercise
covert vector
what u gotta cover and by when
covert vector
you got a course textbook?
sonic current
yeah! but it's in portuguese
covert vector
and does your course have a website and outline
covert vector
do u know portuguese
covert vector
so hopefully u know your own language
sonic current
just pick something that's not too silly so you can build some confidence
you can do the ones with solutions just fine
covert vector
have u tried looking at other group theory notes and books
it's not like this guy's notes are the only resource that exists
covert vector
what book did jichael recommend
if you're struggling a lot, then Gallian is a pretty nice book
has loads of examples and takes its time
sonic current
cute motivational quotes on every chapter which is always a bonus
covert vector
that's 1 of the reasons I hate Gallian lol
sonic current
oh come on some of those are great
there's one I cherish so much
but it's been dropped on the pdf I have, I've only got it in the physical version
check gallian
you can prove the subgroup generated by one element a is a subgroup
that will pave the way for cyclic subgroups and it's a very very very very trivial proof
no, ofc
come on, that wasn't even trying
point is, you can have subgroups that are generated by one element a
so that all elements are powers a^1, a^2, a^3, a^4 etc.
yeah!
you've been there
so, you can prove <a> (notation for a generated subset) is a subgroup
very easy
yes
but you could prove its always a subgroup
and it's a nice proof
stone fulcrum
Think group axioms
stone fulcrum
Yeah I was watching that was intense
stone fulcrum
Let's say a is in a group
Then a² is in the group cuz closure lol
But then a³ also
Every power of a must be in the group.
Furthermore, there must also be a point where aⁿ = e cuz finite
Did we say finite?
I'm saying finite
So <a> is closed, has identity
sonic current
are you done?
record time
too bad half of it is wrong
just kidding, you're good nice work 
just try to be more rigorous when you write it down
algebra benefits largely from rigour
sonic current
it's just a matter of stating clearly what you want
and doing rigorous algebra work to prove something is true
"because anything in <a>
is made of a
and take something else in <a>
it is also made of a
then operating these 2 things together
is still a multiple of a
so it's made of a"
this doesn't fly in a test
you must do something like
Let x,y be in <a>.
Then, x=a^b and y=a^c for integers b,c
a^b*a^c = a^(b+c) which is an element of <a>, because b+c is also an integer
if you must, you can add your trademark at the end of the proof to signify QED
which is, ofc,
¯_(ツ)_/¯
i bet your teacher will find it lovely
sick acorn
square!!
and yeah
i’d prob write... Then xy = (a^b)(a^c) = a^(b+c) in <a> because bc in Z.
sonic current
ah yeah, my bad, it's b+c
sonic current
you did it
just don't slack the exercise
because if you make those general statements, you're bound to be misunderstood
and it simply won't fly with your teacher if you seem not to be comfortable with doing very simple math to prove your point
i don't want to be mean but it's better if you get that habit now, to be rigorous and clear, than in the latter stages where a lack of very clear understanding of the basic concepts will lead you to further confusion
i'll be around if you need more help
delicate chasm
Rip
sonic current
do it
I gotta study a bit but I'm keeping you at a glance
ah that is a fun one
I'm proud that I came up with something close on my own
i think it's my greatest achievement
in math
so far
oblique river
@chilly ocean just a comment on your last proof: you shouldn't make it "more symboly"
symbols are useful because they allow us to communicate complicated ideas quickly. it's like shorthand
but if you use too much shorthand, it becomes very hard to parse
math should be easy to read
and overloading it with notation makes it harder to read
what do you mean?
I never use the \forall symbol when I'm writing something for other people
I only use it in my personal notes/scratchwork
delicate chasm
Symbols allow for conciseness which makes things clearer sometimes 
oblique river
^this
delicate chasm
When you aren't used to the symbols it tends to confuse the matter :P
sick acorn
writing out for all is considered good practice for proofs i think
delicate chasm
It is good to practice writing things down clearly in symbols though
sick acorn
it's fine to use it it for informal notes hahaha
oblique river
@chilly ocean what did I tell you about putting your Gs and Hs and ns in math mode!
the h and n and e should also be in math mode
There is a smallest $n \in \mathbb{N}$ such that for some $h \in H$, $h^n = e$.
cloud walrusBOT
Buncho Bananas:
oblique river
is how the last line should look
cloud walrusBOT
creamy shits:
sonic current
that is very clean
see? i'm proud
specially the e^n step I think you really appreciated that lol
oak perch
hmm so you can use the subgroup test on your test 😃
oblique river
so the argument is right
but you should at some point justify that "the order of a divides n" is equivalent to "a^n = e"
what? it does show that ab in H because (ab)^n = e so the order of ab divides n
still: >but you should at some point justify that "the order of a divides n" is equivalent to "a^n = e"
simple valley
that doesn't look right
that allows multiple n for a single a
the order of an element is the smallest positive n satisfying that condition
cloud walrusBOT
creamy shits:
simple valley
$n \in \mathbb{N}+ \land a^n = e \land \forall m \in \mathbb{N}+, m < n \implies a^m \ne e$
cloud walrusBOT
mniip:
simple valley
obviously if $a^n=e$ then $a^{nk}=e$
cloud walrusBOT
mniip:
simple valley
so if $|a|k = n$ ($|a|$ divides $n$) then $a^n = e$
cloud walrusBOT
mniip:
simple valley
the converse is less trivial, but also true
requires a bit of number theory/cyclic group properties
covert vector
ok.
simple valley
if $a^n=e$ and $|a|=m$ then $n$ can be uniquely decomposed as $n=qm+r$ where $0 \le r < m$, hence $a^{qm+r} = a^{qm}a^r=a^r=e$. If $r \ne 0$ then $|a| = r$ which is a contradiction ($r < m$). hence $n=qm$ and m divides n
cloud walrusBOT
mniip:
covert vector
ADOPT MY NOTATION !! @chilly ocean
at the very beginning, before proving anything, you need to pick an n and stay with it
you can't change n after that
so in your identity section, you can't have that line "there exists a smallest n"
you can't say n anymore cuz that's already reserved for the subset H_n
otherwise ur reusing variables
covert vector
what do you mean by let h^m = e
can we pick specific numbers and try to parse stuff
let's say n = 4
so we have the subset H_4 of elements g in G such that g⁴ = e
what does this tell us about the order of g
(note: it does not tell us that it's 4)
Ya
so what are the possible orders of g
given that g⁴ = e
A HUNDRED
oh wait nvm
Ok
Ya so what's the order of e in G
ok cool, that's part of 1,2,4
so it passes, e is in H_4
now u wanna show that if x is in H_4
so is its inverse
so what do we start with
Ok
Ok
what can we conclude from x is in H_4
Ya
Ya u can say like suppose x has order m, where m divides n (4 here )
I mean it's mostly fine here,
except at some point u changed from x to h
and you should specify where the element is taken from
cloud walrusBOT
creamy shits:
covert vector
lol
sick acorn
for the identity portion
why not write it as like
Let $h \in H_n$. Then $e = h^n \in H_n$.
cloud walrusBOT
Vic:
ocean marsh
How do you know that H_n is nonempty?
sick acorn
wouldn't the identity in G have order 1? And 1 divides n?
ocean marsh
Since the identity is the only element guaranteed to be in there for an arbitrary group and integer, that was reallly circular
sick acorn
yeeeeeeep whoops
cloud walrusBOT
creamy shits:
delicate chasm
\sres{If an element $a$ of a group $G$ satisfies $a^m = e$, then the order of $a$ divides $m$.}
\begin{proof}
Let $a\in G$ have order $n$, and assume that $a^m = e$.\
Since $n$ is the smallest natural number such that $a^n = e$, we know that $n \leq m$.\
By the division remainder theorem there exists $k,r \in \bN$ with $ 0 \leq r < n$ such that:
$$m = kn + r.$$
Our goal is to prove that $r = 0$.\
Now, we have
$$e = a^m = a^{kn+r} = a^{kn}a^r = (a^n)^k a^r = ea^r = a^r.$$
Hence $a^r = e$. However, $0 \leq r < n$ and $n$ is the smallest natural number such that $a^n = e$.\
Hence we must have $r = 0$.\
Thus $m$ can be written as $m = kn$ and we have that $n$ divides $m$, as desired.
\end{proof}
cloud walrusBOT
Puerøsola:
delicate chasm
@chilly ocean
simple valley
is this still going
delicate chasm
Nearly done
simple valley
@delicate chasm ooh that looks exactly like what I wrote earlier :P https://discordapp.com/channels/268882317391429632/496784958430380033/528056700305932301
delicate chasm
LOL
I wasn't here for that portion of the saga
Okay, so now what you have shown in various places is that $a^m = e$
cloud walrusBOT
Puerøsola:
simple valley
you want the converse statement too
sonic current
have you gone to sleep yet, @chilly ocean?
simple valley
if order of a divides m then a^m=e
sonic current
okay
sonic current
I went to sleep in the meantime
simple valley
also, you're doing some weird stuff to prove that the subset you're talking about is a subgroup
delicate chasm
There's an easier check, sure, but this might be best for understanding
simple valley
you don't need to check the inverse laws or anything, only that the identity is in the subgroup, and that it's closed under multiplication and inverses
equational laws are all satisfied by the "ambient" group
delicate chasm
They aren't checking the laws as far as I can see?
sonic current
what's the problem right now?
simple valley
I can't for the life of me figure what happens in the "Inverse:" section
delicate chasm
The inverse section is okay apart from the last step being missing
They eventually show that $(h^{-1})^m = e$
cloud walrusBOT
Puerøsola:
delicate chasm
which, along with the result from before, gives what we want
Everything in the inverse section is for that purpose
simple valley
you need to clean up your... sentences
delicate chasm
Incidentally, that proof doesn't show that the order of $h^{-1}$ is $m$, it shows that the order of $h^{-1}$ divides $m$
cloud walrusBOT
Puerøsola:
errant drum
It could be a bit more explicit in saying the order is m
errant drum
Also how do you get the white background on Latex @delicate chasm
errant drum
,tex --colour white
cloud walrusBOT
Your colour scheme has been changed to white
errant drum
Thanks
delicate chasm
delicate chasm
You just need to show the order divides n
So your goal is to show the order of $h^{-1}$ divides $n$
cloud walrusBOT
Puerøsola:
delicate chasm
What you have shown is that if the order of $h$ is $m$, then the order of $h^{-1}$ divides $m$
cloud walrusBOT
Puerøsola:
delicate chasm
So, if the order of $h$ divides $n$, then the order of $h^{-1}$ divides the order of $h$ and hence divides $n$.
cloud walrusBOT
Puerøsola:
simple valley
a divides b $\iff \exists c : ac = b$
delicate chasm
Oof
cloud walrusBOT
mniip:
errant drum
Yeah
delicate chasm
Yeah what I wrote is correct I think
errant drum
Oh i see
cloud walrusBOT
creamy shits:
simple valley
why are "Inverse" and "Closure" two different sections
errant drum
They are different axioms @simple valley
simple valley
yeah
errant drum
Although it feels weird not seeing closure first
delicate chasm
While you can combine them you might as well do it one step at a time initially
simple valley
@errant drum I would think that closure subsumes both closure under inverse and closure under multiplication
also arguably closure under the nullary "identity" operation
cloud walrusBOT
creamy shits:
delicate chasm
@chilly ocean When you're happy with your proof, give me a ping and I can maybe show you a slightly cleaner way of formatting it
errant drum
No they have the same order
delicate chasm
We haven't shown they have the same order though
(although that is true)
Slightly, yes
errant drum
Your final piece of the proof for inverse should be (h^-1)^m = e
simple valley
are you familiar with the $(ab)^{-1} = b^{-1}a^{-1}$ law
cloud walrusBOT
mniip:
simple valley
you keep trying to go via the definition of the inverse, while that's totally unnecessary
you can just equate $(h^m)^{-1}$ with $(h^{-1})^m$
cloud walrusBOT
mniip:
delicate chasm
That's what they did, just roundabout
But the issue atm is not that it is a roundabout method, it is that it is incomplete
errant drum
@chilly ocean It's not explicitly shown
delicate chasm
They explicitly wrote down that $e = (h^{-1})^m$
cloud walrusBOT
Puerøsola:
delicate chasm
With half a line more, it's done
One sentence really
@chilly ocean Okay, follow me for a sec
You showed that $e = (h^{-1})^m$, correct?
cloud walrusBOT
Puerøsola:
errant drum
It's not stated anywhere in that proof
errant drum
Oh I see
delicate chasm
Okay, so you have jumped the gun a little in the next point and said that $|h^{-1}| = |h|$
cloud walrusBOT
Puerøsola:
delicate chasm
Note: This is true.. however you haven't proved it quite yet
Do you remember the result we did before?
If $a^m = e $ then $|a| \mid m$
cloud walrusBOT
Puerøsola:
cloud walrusBOT
Puerøsola:
delicate chasm
Hence by the result, $|h^{-1}| \mid (m = |h|)$
cloud walrusBOT
Puerøsola:
delicate chasm
By assumption we have that $|h| \mid n$
cloud walrusBOT
Puerøsola:
delicate chasm
Hence $|h^{-1}| \mid |h| \mid n$ and so $|h^{-1}| \mid n$.
cloud walrusBOT
Puerøsola:
delicate chasm
Yes, that's a good enough argument
simple valley
an even better argument would be
errant drum
Don't write |h^-1| = |h|
delicate chasm
We will get to that
simple valley
Suppose $h \in H_m$, therefore $|h| \mid m$. By lemma 1, $h^m = e$, therefore $(h^{-1})^m = e$, and by lemma 1, $|h^{-1}| \mid m$ and hence $h^{-1} \in H_m$
cloud walrusBOT
mniip:
delicate chasm
Yeah, but we are working with what is there for now
simple valley
all it takes
delicate chasm
Not re-writing everything
simple valley
shrug
cloud walrusBOT
creamy shits:
errant drum
For closure you've assumed a & b have the same order.
So suppose a has order m and b has order k
a^m = e, b^k = e
Now because both k and m divide n
n = cm and n=dk for some integers c,d
Using the stuff above what can you say about ab
Remember to prove closure you need to show whatever the order of ab is it must divide n
cloud walrusBOT
creamy shits:
simple valley
what can you say about (ab)^n
sonic current
back to my cave
delicate chasm
Where are we at now?
errant drum
The end result is to show that ab has an order that divides n
@delicate chasm The closure part of the proof he assumed that the order of the two elements was the same
delicate chasm
Oof
simple valley
@chilly ocean "start with conclustions" (ab)^n is just an expression
it's not a statement
that's wrong
delicate chasm
Yeah
cloud walrusBOT
creamy shits:
delicate chasm
That doesn't make sense, crem
Why not
If you show that $(ab)^n = e$ then you are done, right?
cloud walrusBOT
Puerøsola:
errant drum
We're just calculating (ab)^n. If it works it works if it doesn't we'll try something else
delicate chasm
So just manipulate $(ab)^n$ until you get to $e$
cloud walrusBOT
Puerøsola:
delicate chasm
It's a matter of turning the LHS into the RHS
errant drum
Yes
delicate chasm
Standard algebraic manipulation, essentially
errant drum
The answer is (ab)^n = e
We just told you to work with (ab)^n
That's not the same thing
delicate chasm
If I asked you to show that $(a+b)(a-b) = a^2 - b^2$
cloud walrusBOT
Puerøsola:
delicate chasm
You would presumably start with the LHS and turn it into the RHS
This is the same, you are asked to show that $(ab)^n =e$
cloud walrusBOT
Puerøsola:
simple valley
the "answer" is x=y
to prove the "answer" you can start with either x or y
doesn't matter
cloud walrusBOT
creamy shits:
delicate chasm
Yes
errant drum
Yeah that works. But explain why a^n = e and b^n =e
And that's your proof done
Yeah
If you wanted to say it more mathematically:
Suppose a^m = e, b^k = e.
Then because m|n
n=cm for some natural number c.
so a^n = a^cm
a^cm = (a^m)^c = e^c = e
And same for b
cloud walrusBOT
creamy shits:
errant drum
How does one even catch that
errant drum
You're a robot sent from the future to destroy us and I won't accept any other theory @delicate chasm
delicate chasm
Being all okay requires more than being mathematically correct
Lmao
So
There are a few things that can be massively simplified
And the formatting needs some work
Now that you are done
want me to show you my version?
Using the same outline
delicate chasm
We will use $|a|$ to denote the order of an element $a$ of a group $G$.
\sres{Let $a$ is an element of a group $G$. Then $a^n = e$ if and only if $|a| \mid n$.}
\sres{In an abelian group $G$, the set of elements whose order divide a fixed integer $n$ is a subgroup.}
\begin{proof}
Let $H$ denote the subset of $G$ consisting of elements whose order divide $n$. We shall show that $H$ is a subgroup of $G$.\
We do this by showing existence of identity, closure under multiplication, and closure under inverses:
\begin{itemize}
\item Existence of identity. To show: $e \in H$.\
The order of $e$ in $G$ is $1$, since $e^1 = e$. Since $1$ divides $n$, $e \in H$ as desired.
\item Closure under multiplication. To show: If $a,b \in H$ then $ab \in H$.\
By Result 1, it suffices to show that $(ab)^n = e$.\
To see this, we note that since $G$ is abelian, we have $(ab)^n = a^nb^n$. Again using Result 1, we have:
$$(ab)^n = a^nb^n = ee = e.$$
Hence the order of $ab$ divides $n$, and $ab \in H$ as desired.
\item Closure under inverses. To show: If $a \in H$ then $a^{-1} \in H$.\
Again by Result 1, it suffices to show that $(a^{-1})^n = e$. To see this, we have:
$$(a^{-1})^n = a^{-1\cdot n} = (a^n)^{-1} = e^{-1} = e$$
where we again used Result 1 to say that $a^n = e$.\
Hence the order of $a^{-1}$ divides $n$ and we have $a^{-1} \in H$, as desired.
\end{itemize}
Since $H$ is a subset of $G$ with identity, closed under multiplication and inverses, it must be a subgroup of $G$, and we are done.
\end{proof}
oblique river
it looks mathematically correct but the wording could be improved
cloud walrusBOT
Puerøsola:
oblique river
@delicate chasm should you "prove" result 1 or did you decide that was obvious enough to omit?
simple valley
we proved it multiple times above
oblique river
yes but it's not in pur's writeup
delicate chasm
I'll add it
oblique river
I'm asking specifically about pur's exposition
errant drum
Look how pretty that formatting is
oblique river
it's also easy to read
not overrelying on symbols
only includes relevant details
I approve 👍
simple valley
in my mind I would represent it a bit differently
we need to prove $a, b \in H \implies ab \in H$. By definition of $H$ that is $|a| \mid n \land |b| \mid n \implies |ab| \mid n$. By result 1 that is $a^n = e \land b^n = e \implies (ab)^n = e$
cloud walrusBOT
mniip:
delicate chasm
We will use $|a|$ to denote the order of an element $a$ of a group $G$.
\sres{Let $a$ is an element of a group $G$. Then $a^m = e$ if and only if $|a| \mid m$.}
\begin{proof}
As is usual for an if and only if proof, we first prove one way then the other.
\begin{itemize}
\item First, we show that if $a^m = e$ then $|a| \mid m$.\
Let $n = |a|$.\
$n$ is the smallest natural number such that $a^{n} = e$, so we have $n \leq m$.\
Hence there exists some $k,r \in \bN$ with $0\leq r < n$ such that $m = kn +r$.\
We wish to show that the remainder $r$ is $0$.\
Now, we have
$$e = a^m = a^{kn+r} = a^{kn}a^r = (a^n)^k a^r = ea^r = a^r.$$
Hence $a^r = e$. However, $0 \leq r < n$ and $n$ is the smallest natural number such that $a^n = e$.\
Hence we must have $r = 0$.\
Thus $m$ can be written as $m = kn$ and we have that $n$ divides $m$, as desired.
\item Next we show that if $|a| \mid n$ then $a^n = e$.\
To see this, since $|a| \mid n$ we can write $n = k |a|$ for some $k \in \bN$.\
Hence $$a^n = a^{k|a|} = (a^{|a|})^k = e^k= e$$ as desired.
\end{itemize}
\end{proof}
\sres{In an abelian group $G$, the set of elements whose order divide a fixed integer $n$ is a subgroup.}
\begin{proof}
Let $H$ denote the subset of $G$ consisting of elements whose order divide $n$. We shall show that $H$ is a subgroup of $G$.\
We do this by showing existence of identity, closure under multiplication, and closure under inverses:
\begin{itemize}
\item Existence of identity. To show: $e \in H$.\
The order of $e$ in $G$ is $1$, since $e^1 = e$. Since $1$ divides $n$, $e \in H$ as desired.
\item Closure under multiplication. To show: If $a,b \in H$ then $ab \in H$.\
By Result 1, it suffices to show that $(ab)^n = e$.\
To see this, we note that since $G$ is abelian, we have $(ab)^n = a^nb^n$. Again using Result 1, we have:
$$(ab)^n = a^nb^n = ee = e.$$
Hence the order of $ab$ divides $n$, and $ab \in H$ as desired.
\item Closure under inverses. To show: If $a \in H$ then $a^{-1} \in H$.\
Again by Result 1, it suffices to show that $(a^{-1})^n = e$. To see this, we have:
$$(a^{-1})^n = a^{-1\cdot n} = (a^n)^{-1} = e^{-1} = e$$
where we again used Result 1 to say that $a^n = e$.\
Hence the order of $a^{-1}$ divides $n$ and we have $a^{-1} \in H$, as desired.
\end{itemize}
Since $H$ is a subset of $G$ with identity, closed under multiplication and inverses, it must be a subgroup of $G$, and we are done.
\end{proof}.
simple valley
well, note how it took each of us only a few moments to come up with a proof
it comes with experience I guess
delicate chasm
The first result is supposed to be "obvious", and all the parts of the second proof follow almost immediately from the first result
simple valley
proof writing is its own discipline
delicate chasm
Aye, you simply haven't had enough exposure to these concepts
simple valley
maybe it isn't the group theory that's the problem here but rather mathematical logic in general?
delicate chasm
Have you done a discrete maths course with proof, crem?
Hmm
Have you done any discrete maths course?
simple valley
the opposite of analysis?
simple valley
graph theory, combinatorics, boolean algebra, etc
simple valley
sounds like you have no experience writing formal proofs whatsoever
errant drum
Does your university have an introduction to abstract maths course or a similar kind of course?
Or a foundations of maths course?
Or any course with that kind of name
Yeah
delicate chasm
A first year discrete maths/ mathematcial proof course usually introduces you to the basics of how to write a formal proof, and how to manipulate the basic atomic concepts. It also revises things like the division remainder theorem, fundamental tools in our every day toolbox
sonic current
yeah, discrete maths is important
oblique river
@chilly ocean I think the quesiton is: have you ever written a proof before this course
delicate chasm
^
errant drum
If you can try and read the textbook for that foundations course
oblique river
is that is?
errant drum
That's different
simple valley
that's not a formal proof really
sonic current
nope
oblique river
well in that case your struggles are probably not as much with the material but really with how to write a proof
sonic current
my discrete math course had modulo theory, prime numbers, euclidean division, etc.
errant drum
^
errant drum
Modular arithmetic
delicate chasm
That's what the course I tutored was like as well, protsac
oblique river
but no proofs unfortunately
sonic current
and I had another one very early on where I did some very basic proofs, like √2 is irrational etc.
delicate chasm
Can't remember whether I did the same sort of course
errant drum
What was your first uni maths course @delicate chasm
simple valley
my first formal proofs happened to be in a first year analysis course, but by that point I was already familiar enough with curry-howard to instantly grok what's going on
delicate chasm
Nice
errant drum
It's really important that your understanding of that foundations course is 100% solid @chilly ocean
delicate chasm
Curry howard is pretty up there
Different way of thinking though
You might be interested in a seminar series that happened last year at my uni
simple valley
is it really different
to prove a forall-statement you usually write "fix x", p much the same as writing "lambda x. "
delicate chasm
oblique river
@chilly ocean I think what this all comes down to is that writing proofs really just comes with experience
and right now, you just don't have that experience. that's not saying you're dumb, it's just that you literally haven't had a real proof-based math course before this one
so it's going to be difficult for you. it's not standard (at least in american universities) for group theory to be a student's first introduction to proof-based math
simple valley
@delicate chasm the best part was when in the formal systems course there were exercises where you have to prove a certain statement without using the deduction theorem
just convert a lambda term into a SKI term and you're done!
oblique river
all i'm saying is that you're not dumb, you're just inexperienced, and the only way to get that experience is to try to expose yourself to more proofs and to try to write more proofs
delicate chasm
I have no experience with lambda calculus beyond bedtime reading :P I'm a cat theory/ homological algebra person
Good luck crem
oblique river
fast enough for what?
delicate chasm
It's going to be a steep climb for a bit
errant drum
Real analysis feels like that
simple valley
@delicate chasm well are you familiar with the internal lambda calculus of a category?
oblique river
well you're going to have to learn to write them yourself
simple valley
I think a cartesian closed category is sufficient?
oblique river
have you talked to your professor?
what have they said?
do they know that you struggle with writing proofs? or do they think you're just struggling with the material?
errant drum
Well then practice but make some space to go through that foundations course later on
chilly ocean
Maybe some book like "How to prove it" could help you 
simple valley
oh yeah "How to prove it" is a great book
delicate chasm
Cartesian closed cat is equivalent to having a model for a typed lambda calculus?
Yah it is
I grew up on it
oblique river
well that sounds like a shitty prof if that's all they say
chilly ocean
It's probably on libgen so you can save some money if ebooks are okay for you 
errant drum
I mean the course structure is off
chilly ocean
What semester are you in?
simple valley
@delicate chasm you need a couple tricks to encode lambda calculus terms in terms of eval and various monoidal category shenanigans but other than that it's the same
well, eval and the associator really
delicate chasm
associator! something I understand
simple valley
$eval : B^A \times A \to B$
cloud walrusBOT
mniip:
delicate chasm
Looks like an evaluation morphism of sorts
evaluate the first of pair on the second?
simple valley
apply the left to the right yes
simple valley
now if you have a morphism $h : A -> B$, you could use the unitor to transform it into $1 \times A \to B$
cloud walrusBOT
mniip:
simple valley
and then by the universal property of the exponential there's a morphism $1 \to B^A$ that commutes with h
cloud walrusBOT
mniip:
simple valley
so in effect if you have $h : A \to B$ you can turn it into a "point" in $B^A$
cloud walrusBOT
mniip:
delicate chasm
Yep
simple valley
lambda calculus!
delicate chasm
Well it's a 1-morphism in the 2-cat already
simple valley
actually
just use the adjunction $\operatorname{Hom}(1 \times A, B) \cong \operatorname{Hom}(1, B^A)$
cloud walrusBOT
mniip:
delicate chasm
Exponential adjunction
simple valley
maybe this is best moved into aanother channel
delicate chasm
$(B)^{1 \times A} \isom \big(B^A\big)^1$
cloud walrusBOT
Puerøsola:
cloud walrusBOT
The current time for Pur is 3:26 AM (AEST(+1000)) on Sat, 29/12/2018
delicate chasm
I should sleep though
Did you understand the proof I posted before?
Do you have a list of helpers and the number of hours from each? xD
simple valley
well that's just the previous exercise with n=infintity!
delicate chasm
Lmao
sonic current
I wish I had this stamina for my own exams jesus
delicate chasm
Same
simple valley
is this a straightforward application of the second iso thm or am I missing something?
$HK/K \cong H/H \cap K \implies #(HK/K) = #(H/H \cap K) \implies #(HK)/#(K) = #(H)/#(H \cap K)$
cloud walrusBOT
mniip:
stone fulcrum
What does # mean?
stone fulcrum
And of course, the torsion are the elements that have finite order
cloud walrusBOT
creamy shits:
raw moth
you don't need to also check the existence of an identity
closure under multiplication and existence of inverses also gives you existence of an identity
raw moth
ya ok fair
simple valley
if x in H then xx^-1 = e in H
stone fulcrum
It's trivial, since e¹ = e, but worth stating
simple valley
but considering the torsion part of a group can be just the identity
it's best to just prove it has an identity instead
what
cloud walrusBOT
creamy shits:
simple valley
why the "as $\forall t\in T$ , $t\cdot 1 =1\cdot t=t$"
cloud walrusBOT
mniip:
simple valley
it has no relevance to the matter
so?
well why is it 1?
definitely not because et=t
(in practice, yes you can just state that, it's an "obvious" statement, but you should figure out why anyway)
cloud walrusBOT
creamy shits:
simple valley
@chilly ocean what
yes but that fact is irrelevant here
can you write down the definition using quantifiers?
or at least, using words but with some actual variables
(point-free english )
raw moth
n positive
cloud walrusBOT
creamy shits:
raw moth
right, so the order of the element 1 is 1 because?
yes, and why?
I assumed you were using 1 to denote the identity of the group G
o ok
in which case
simple valley
ok you've stated
$\exists$ smallest $n\in \mathbb N$ s.t $a^n=e$ for some $a\in G$
cloud walrusBOT
mniip:
raw moth
you want to find the order of the element e then
raw moth
you said you were calling it e... doesn't matter, but whatever you are calling the identity, show that it's order is 1!
by using the definition of order
simple valley
don't rush, one step at a time
you substituted t for a
I asked to substitute the identity for a
because that's what we're trying to look at
why not?
no
simple valley
@chilly ocean do you know what instantiation is
raw moth
yes we're just trying to get you to justify that the order of the identity in G is finite (specifically, 1)
simple valley
when you know that $\forall x, P(x)$, you are allowed to "instantiate" x to any value, by substituting that value instead of x: $P(value)$
cloud walrusBOT
mniip:
simple valley
a proper definition of an order of an element would be
raw moth
let t be the identity of G
then the order of t is the smallest positive integer n such that t^n = t
you know it has an identity
and you're calling it t
simple valley
$\forall a, \forall n,$ "n is the order of a" $\iff$ n is the smallest number such that $a^n = e$
cloud walrusBOT
mniip:
simple valley
in particular, picking a=e,
$\forall n,$ "n is the order of e" $\iff$ n is the smallest number such that $e^n = e$
cloud walrusBOT
mniip:
simple valley
clearly the only value of n that satisfies this formula is 1
well rather
the only value of n that satisfies the right hand side
therefore we conclude "1 is the order of e"
that's what I meant by substituting e in place of a
if a statement holds in general, it holds in a specific case too
that's the general definition
what
no
identity is an element of your group
n is a number
not an element of your group
you said n=1 is the identity
also you're still interpreting what I said backwards
I'm not saying t^1 = e hence t=e. I'm saying smallest n such that e^n = e is n=1.
what
"t is the identity if there is a smallest n such that t^n=e"
that's just wrong
in a finite group, all elements satisfy that condition
but most of them aren't the identity
is all of this group theory a class you're required to take or just doing some extracurricular reading?
without elementary logic you're in for a tough time
my suggestion is to put this on hold and learn some mathematical logic and proof writing
well that's a problem then
errant drum
How far away is your exam?
Is it part of your grade or just a mock exam?
I see
Well RIP everyone
simple valley
heh reminds me of my PLT exams
in the definition of order you have a^n=e
you literally
substitute
a for e
because you are asked the order of e
sub e for a
I don't know how to explain
covert vector
you need to review mathematical logic and proof writing
simple valley
I can't figure out what kind of snag you're hitting
covert vector
I don't care if you only have like 2 weeks, this is a pretty massive gap
and 2 weeks is actually a reasonable amount of time 🤔
wot exams ya got
wat else
errant drum
Do you have more proof based exams?
covert vector
🤔
errant drum
I mean do you have any other proof based exams in Jan?
I'd say you should probably review logic properly but since you don't have much time read chapters 5-8 of this book
You can read it in a day
errant drum
Hmmm I just clicked on the first link I saw
If that book doesn't work look for a PDF of a book called "how to prove it"
simple valley
of "How to prove it" you really only need the first three chapters
simple valley
does the third iso thm imply that composition of two canonical maps is a canonical map?
that is, if $N, K \lhd G$, $N \le K \le G$ then we have two canonical maps: $G \to G/N$ and $G/N \to (G/N)/(K/N) \cong G/K$
cloud walrusBOT
mniip:
simple valley
is their composition equal to the canonical map $G \to G/K$?
cloud walrusBOT
mniip:
simple valley
I guess yes, $x \mapsto xN \mapsto xK$
cloud walrusBOT
mniip:
simple valley
that's evident
the question whether composition of these two homs is this exact hom
simple valley
okay so in that case, not only a morphism f : G -> H factors through G/ker f
but also through any G/N where N is a normal subgroup of ker f
cloud walrusBOT
creamy shits:
simple valley
oh god
lyric falcon
but you know the identity has order 1
that's like the definition of identity
$e^1=e$
cloud walrusBOT
Gonzo17:
simple valley
that's backwards
you're not supposed to say t^n = t
an element e such that forall x, ex = xe = x
there's a theorem that there is only one identity in a group
which is why we call it the element e
it's a very concrete element
in the definition of "order" we simply refer to it by the name "e"
jeez
no
you just substitute into the definition
cloud walrusBOT
creamy shits:
simple valley
the fuck is t
cloud walrusBOT
creamy shits:
cloud walrusBOT
creamy shits:
simple valley
it's not an unknown t either
you're doing the reverse
you have an element
and you're trying to see what is its order
actually put off your task for a second
consider $C_6 = {0, 1, 2, 3, 4, 5}$
cloud walrusBOT
mniip:
simple valley
what is the order of the element 4?
why?
?
what about the definition of order?
for some reason
you have no trouble substituting 4 for a here
you have a concrete element 4
similarly, you have a concrete element e in T
in fact, C_6 is abelian, so T might by chance be C_6
cloud walrusBOT
creamy shits:
cloud walrusBOT
creamy shits:
sonic current
i'm with you, buddy
cloud walrusBOT
creamy shits:
simple valley
that's still a non-sequitur
cloud walrusBOT
creamy shits:
simple valley
use the definition
of order
"let n be the order of a, that is, the smallest number such that a^n=e"
cloud walrusBOT
creamy shits:
simple valley
"Then |ab|=m" that's false
cloud walrusBOT
creamy shits:
simple valley
it might not be
cloud walrusBOT
creamy shits:
simple valley
you keep chanting the same
a^n b^m = e but that doesn't say anything about ab
abandon a^n b^m
consider (ab)^(nm)
and what does that tell us
cloud walrusBOT
creamy shits:
cloud walrusBOT
creamy shits:
simple valley
(ab)^(nm)=e doesn't imply |ab|=nm
it doesn't
yes
but if you just have a^n=e that doesn't mean |a|=n
we kind of touched this yesterday but I suspect it might have gone completely over your head
cloud walrusBOT
creamy shits:
simple valley
it might not be, yes
but there's a weaker statement we can make instead
sigh
well I guess that works
yesterday we discussed how a^n=e implies that |a| divides n
but "a^n = e implies |a| is not more than n" is fine too
that's a different n
"q^z = e implies |q| is not more than z"
there's no "biggest n"
you can multiply the order by an arbitrary number
no
sonic current
i'd just rephrase the let |a|=n etc.
to let |a| = n, |b| = m, n m smallest n,m s.t.
ah nevermind you're not trying to apply it to both are you?
don't mind me
it's done
now inverses
simple valley
well first wrap this into an exponent
sonic current
yeah I'd do that too
change the notation
then find the exponent n such that z^n = 1
civic linden
into re^itheta?
simple valley
yes
civic linden
so Z^n=1 implies that r^n=1 right?
simple valley
huh?
civic linden
so i got z^n=r^ne^in2pi/3
simple valley
and r?
simple valley
arctan you mean
civic linden
ye
simple valley
well do you know the formula for r?
simple valley
well? r=?
civic linden
I have a question, we have G=GL(2,ℝ) where G is a group.
how do I work out what <x> is ?
say the matrix consisting of [cospi/4...sinpi/4...-sinpi/4... cospi/4]
worthy kindle
I suppose the subgroup generated by x is the set made of the powers of x and of the powers of the inverse of x,
civic linden
yes
worthy kindle
so you'd have to find a nice formula for [your matrix]^n and for [the inverse of your matrix] ^n
worthy kindle
What do you mean?
worthy kindle
A^0 = I
civic linden
yah so I just have to find the others right?
worthy kindle
Yea
sullen flint
finding powers of matrices is easy when you can diagonalize them
bleak abyss
Well in this case you should note a geometric description of the linear map this matrix represents
That'll make your life easier because instead of multiplying matrices you can just geometrically interpret what it means to compose that operation and easily find the matrix form
cloud walrusBOT
creamy shits:
lyric falcon
it's complex numbers with multiplication?
How is the identity 2pi ?
The identity would be 1
you might mean that it's $e^{2\pi i}$
errant drum
That's cos(2pi) + isin(2pi)
cloud walrusBOT
Gonzo17:
cloud walrusBOT
creamy shits:
loud crystal
can i PM any1 and try to explain my <probbably simple> engineering level maths question? I can't really translate it into english directly so not sure if i should spam here
basic linear function tho
errant drum
@loud crystal post it in a questions channel
lyric falcon
tbh when you're dealing with multiplication in C you usually want to use the trigonometric form
do you know what it is?
cloud walrusBOT
Cosmicrays:
errant drum
Yeah
Yeah when you deal with complex functions you choose a range of angles you want to work in typically
-pi to pi or 0 to 2pi
Yeah
lyric falcon
yeah it's not all of radius 1
you already know you can represent your numbers as $e^{ix}$ for some $x$
cloud walrusBOT
Gonzo17:
lyric falcon
and you want such numbers that $z^n=1$ for some n
cloud walrusBOT
Gonzo17:
cloud walrusBOT
creamy shits:
lyric falcon
But it's not the correct answer
As I said the condition is $z^n=1$ for some $n$ right?
cloud walrusBOT
Gonzo17: