help-33

[kind iris]

so you need to check before that if(S>=P) return 1?

okay, that looks good thanks

.close

[marsh citrus]

Channel closed

Closed by @kind iris

Use .reopen if this was a mistake.

[kind iris]

.reopen

[marsh citrus]

✅

[kind iris]

Okay let's say there is T=5, S=4, P=5

then 5*4/min(5, 1)*4*3

ceil(20/12) is 2

but i don't think so it is possible in 2 rounds

[hidden nacelle]

do you need a O(1) solution?

[kind iris]

Nope, if you have any other solution, i would be happy 🙂

[marsh citrus]

@kind iris Has your question been resolved?

[marsh citrus]

@kind iris Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[buoyant jetty]

Let $f(n)$ be the sum of the positive integer divisors of $n$. If $n$ is prime and $f(f(n))$ is also prime, then call $n$ a bouncy prime. What is the smallest bouncy prime?

[elfin berry]

938c2cc0dcc05f2b68c4287040cfcf71

[weary karma]

Just try the natural numbers

There is no specific condition we can apply here probably

[lucid zenith]

what can you say about f(n)?

[buoyant jetty]

f(n) = sigma(n)

? idk

[buoyant jetty]

wdym

[weary karma]

I meant try checking if any of the natural numbers satisfy this

Start from 2

That's a method

I can't really think of a condition to apply if the sum of divisors is prime

[still temple]

Here's an intuitive understanding for this Question: Remember every prime no. has 2 factors (1 & itself) so you need to find a prime no. when added with 1 gives a number whose sum of factors is again a prime no. ....if it helps

[buoyant jetty]

if n is prime
sigma(n) = n^0 + n = 1 + n

[lunar heath]

if n is prime then f(n) is just 1 + n
f(n) = 1 + n but we don't know if its a prime

[still temple]

i have an answer

[buoyant jetty]

we need (1+n) to be a prime aswell

[lunar heath]

looking for f(n+1) = prime but n+1 is not necessarily a prime

isnt it

[buoyant jetty]

?

[still temple]

yeah

but whatever n+1 is , it's further factors must add up to a prime no.

[weary karma]

Nope

[buoyant jetty]

sigma(n+1) = x

x has to be prime I just noticed

[buoyant jetty]

yeah

[lunar heath]

yes

we know for certain its n+ 2 + some possible other factors

i mean just by plugging in primes n=3 works i think

[buoyant jetty]

2,3,5,7,11,13

[lunar heath]

and 7 would be smallest

[weary karma]

Yeah

[lunar heath]

or maybe not

[weary karma]

It's 3

[lunar heath]

there could be a combination with lower thing

[weary karma]

n is the 'bouncy prime'

[lunar heath]

f(f(n)) doesnt have to be greater than n necessarily does it?

oh nvm

then its just 3

[weary karma]

Nope

[lunar heath]

thought f(f(n)) was the bouncy prime

[weary karma]

Number theory involves a lot of brute force

[weary karma]

Happens lmao

[lunar heath]

feels wrong though

to just plug in

until find the answer

[trail hamlet]

ya I think a better question would have been "How many bouncy primes exist"

[weary karma]

I had like a month long course on this a long time ago , 70% of the questions had some kind of brute force involved

[mellow crag]

that's cus ur doing boring/bad qs lol

[weary karma]

Even if you derive an expression, for ordered pairs and stuff you had to plug in values

Those were boring though

[mellow crag]

for this particular question, notice that f(n) = n+1 and then just test stuff until it works

unfortunately not very interesting but like the fact that you want the smallest bouncy prime suggests ur gonna have to test small values

[lunar heath]

well you know that the other factors must add up to an even number thats interesting

[mellow crag]

i.e. here, 2 is not bouncy but 3 is

[lunar heath]

n + 2 + other factors = another prime

if n is prime ignoring 2, other factors must add up to an even number

[mellow crag]

it's hard to predict the exact values of f(n) here

[buoyant jetty]

.close

.close

[marsh citrus]

Channel closed

Closed by @buoyant jetty

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[weak wind]

How does this graph even work?

[regal coral]

40-4=36

18-4=14

[outer lodge]

Difference iof the two above

[weak wind]

oh i see it now thanks

i need help solving it

[devout mauve]

what value can w be

[weak wind]

6

[devout mauve]

what value can x be

[weak wind]

34 or 22

[burnt bridge]

Then what values can y be?

[weak wind]

2 or 70

and 14 or 58

[burnt bridge]

Then find the values that z can be and then the sum of the four greatest values

[weak wind]

z will have 8 values

[burnt bridge]

Yes

Add the four greatest ones together

The four lowest ones I assume can be ignored

[weak wind]

.close

[marsh citrus]

Channel closed

Closed by @weak wind

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[weak wind]

please help me on this question someone

[cedar stone]

Idk us currencies, so how many cents is a nickel, a penny, and a dime?

[weak wind]

a nickel is 5 cents

a penny is 1 cent

and a dime is 10 cents

[cedar stone]

So let us say we have a nickels, b pennies, and c dimes

[weak wind]

so (5a+b+10c)/(a+b+c)=7

[cedar stone]

5a+b+10c=7a+7b+7c=30a+6b+6c

right?

This has multiple solutions

[verbal goblet]

4 mutual

[weak wind]

5a+b+10c=30a+6b+60c, not 30a+6b+6c im pretty sure

[hardy slate]

you'll get two simultaneous equations; what are they?

[cedar stone]

6c only

[weak wind]

oh ok

[weak wind]

i don't know

[hardy slate]

you need to use the fact that, if a nickel were replaced by five pennies, the average would drop to 6 cents per coin

so in that case, there would be (a-1) nickels, (b+5) pennies and c dimes

[weak wind]

oh the two equations are

[cedar stone]

Oh, not all nickels are replaced?

[hardy slate]

oh

my instinct is that you're only replacing one

[weak wind]

(5a+b+10c)/(a+b+c)=7 and ((5(a-1)+(b+5)+10c))/((a-1)+(b+5)+c)=6

i think

[burnt bridge]

It does say "a nickel"

[hardy slate]

close

[cedar stone]

That should give a unique solution

[hardy slate]

remember that the total number of coins also changes in the second case

[cedar stone]

You are probably right

[weak wind]

oh yeah mb

[hardy slate]

but yeah now you can go for it

[weak wind]

i just edited it

[hardy slate]

yeah so now just go for it

[marsh citrus]

@weak wind Has your question been resolved?

Channel closed

Closed by @weak wind

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[dawn rain]

by any chance anyone here procficient in circuits?

[fickle slate]

Idk

[next apex]

https://dontasktoask.com/ ^

[dawn rain]

woopsie i forgot

how do i determine whats parallel and whats in series in this circuit

R2 is in paralle to some load this is a thevenins theorem question

[desert dirge]

Id say theyre all in series

[dawn rain]

why though?

Image
R2 gets short circuited
R3 is our load
so lets not look at that
R1 is parallel to R5 and thats what i have labelled as R1 in my hand drawn diagram and R4 being R2 and R6 being R3
oh yeah were trying to solve for thevenins equivalent resistance

this is for context mabye i explained poorly

[cedar stone]

First R4 parralel R3

That is in series with R6

That is parallel to R5

That is in series with R1

That is parallel to R2

[dawn rain]

sorry i forgot to mention R3 is our load resistor

[cedar stone]

That doesn't change parallel/series

A parralel to B, of same voltage, A srs to B if same current, is the rule

[dawn rain]

.close

[marsh citrus]

Channel closed

Closed by @dawn rain

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lapis sinew]

Hello

[lapis sinew]

can I claim this channel

[burnt bridge]

Yessir

[lapis sinew]

ok hello from italy AGAIN hahah

@burnt bridge

GEOGEBRA question: I have y=x on the first quadrant a segment 0A of length 2
then GOING CLOCK WISE I have
on Y=0 a OB segment of length 3
and so on adding a 1

ok?

I have to trace a spiral but I have no clue

can u help me with that

?

[tepid temple]

have any picture of your thing?

[lapis sinew]

No but i can do

[tepid temple]

would be great so someone can help you

[lapis sinew]

you see the spiral clock wise?

[tepid temple]

yes

[lapis sinew]

is logarithmic or archimedean i have no clue but i think archimede's more right?

i m not good at mathematika

[tepid temple]

that's more arithmetic cause your augmenting in regular steps

anyways

[lapis sinew]

ok

[tepid temple]

you are supposed to draw a spiral right?

[lapis sinew]

yes

[tepid temple]

but you have all you need so where is the problem?

[lapis sinew]

no wait i dont know to use any thing like the sliders in geogebra

maybe you know about

?

how can i do?

🙂

[tepid temple]

why do you need a slider?

[lapis sinew]

i don t know-.....

but i cannot draw curves right?

[tepid temple]

i dont think thats what you need to do no

[lapis sinew]

no?

[tepid temple]

just trace a line between each point

should do it

[lapis sinew]

ok sir thank you

for the good suggestion

I go at work

[iron meadow]

just graph r = theta in polar form

[lapis sinew]

theta is 45°

hoping that i don t say a bad thing

i mean i have the y=x y=-x and the axis

[tepid temple]

i think this is to advanced math 😅 thats why i didnt suggest you this

[lapis sinew]

but @tepid temple gave me right idea

thank you again

[tepid temple]

no problem 🙂

[lapis sinew]

bye all

[lean flame]

Hello children, what's pooppin

[lapis sinew]

it was a pleasure!!!!! 🙂

big things

[lean flame]

Weire .close

[lapis sinew]

see you on new york post

😄

[lean flame]

.close to close

[burnt bridge]

Ciao

[lapis sinew]

good luck

ciaooooooo

https://www.youtube.com/watch?v=aq0m0hbCjFQ&list=RD8zuCBaG1m3k&index=7

for you help

THANK YOU

[marsh citrus]

@lapis sinew Has your question been resolved?

[marsh citrus]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warped basalt]

I know how to calculate eatch term, but im wondering if theres a smarter way to do this question, other than calculating multiple terms, then trying to find a pattern

[fathom ridge]

I guess you know that ω is a complex cube root of unity

[warped basalt]

ofc

not a cube root

its a square root here

[fathom ridge]

huh?

[warped basalt]

wait

[stoic slate]

[warped basalt]

ohhh

that yea yea

youre right

its a cube root of unity

[fathom ridge]

can you handle it now?

[warped basalt]

i know one way to get the solution, im asking if there is a smarter way to do it

[fathom ridge]

not sure what you mean by that

you just need to exploit its periodicity

[warped basalt]

the way i got to the answer is by calculating multiple terms, then trying to find patternss

[fathom ridge]

Hmm

[stoic slate]

The pattern is the cycle

[fathom ridge]

yes

[lunar heath]

was thinking about rewriting with e^itheta

[warped basalt]

wouldnt that be more annoying

[stoic slate]

Yes but u wanted smarter way

[warped basalt]

ye it is periodic

[warped basalt]

is it shorter that way?

[stoic slate]

If u show your whole solution

I can say

[warped basalt]

uhh

[stoic slate]

Take a picture

[warped basalt]

tbh, im redoing this question, i know how to get there but i only have a part of the actual solution down, id have to dig through a few notebooks to get the whole thing

[stoic slate]

It is impossible to say if there is a shorter way without seeing the actual way you did

When u have it show it

[warped basalt]

ill show you what i have down then

it is a bit messy tho

[warped basalt]

here

[fathom ridge]

Damn

where does it start

[lunar heath]

pretty sure there was an identity with . + 1/z

[stoic slate]

Not what i was expecting

[warped basalt]

its all over the place, thats why i didnt post it before

[warped basalt]

?

i toldya i was calculating the first few terms

[fathom ridge]

[ \omega^2 + \omega + 1 = 0 \implies \omega + \f 1{\omega} = -1 ]

[elfin berry]

neon

[fathom ridge]

this?

[fathom ridge]

oh

sure

[warped basalt]

im afraid thats out of syllabus

[lunar heath]

yeah i think it is

[warped basalt]

ive never seen that

[fathom ridge]

I mean it's literally Euler's identity

[lunar heath]

it would make life easier though

[fathom ridge]

@warped basalt do you still need guidance?

[warped basalt]

we were barely taught on it tbh

yeye

if you have a shorter way

[lunar heath]

yeah maybe not then

[fathom ridge]

does this make sense?

I divided through by ω

[warped basalt]

yup

[sudden panther]

w^3=1

[lunar heath]

what can you say about other terms

[warped basalt]

also ye

[sudden panther]

every 3 power is 1

[warped basalt]

all powers of 3 become 2

ye

[elfin berry]

neon

[sudden panther]

so w^n+w^(-n)=2

nice

[warped basalt]

not allways tho

[sudden panther]

now this

[warped basalt]

only when its a power of three

[sudden panther]

i meant that

[warped basalt]

ahk

[fathom ridge]

every three parentheticals you get -1*-1*2 =2

[warped basalt]

yeye

[fathom ridge]

33 such

so 2^33

[warped basalt]

ah

[fathom ridge]

then there's the ^100 term remaining

which will be -1

[stoic slate]

So your cycle is

[warped basalt]

nah i just got what you meant by periodic nature

[stoic slate]

-1, -1, 2

[warped basalt]

that makes sense

thanks ❤️

.close

[marsh citrus]

Channel closed

Closed by @warped basalt

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sudden osprey]

I don't understand what group of numbers does U15 represent

[pliant siren]

all positive integers less than 15 that are coprime with 15

[sudden osprey]

Ohhhh

Thanks bro

[pliant siren]

you’re welcome

!done

[marsh citrus]

If you are done with this channel, please mark your problem as solved by typing .close

[sudden osprey]

.close

[marsh citrus]

Channel closed

Closed by @sudden osprey

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[honest creek]

C is wrong lol right?

just want to make sure so that i can appeal for this

[proud basin]

i think its correct?

[honest creek]

y intercept = 5 and x intercept = 0 has a product of 0

[proud basin]

since the product is 0, atleast one of them is 0, and if they are 0 then they pass trough the origin no?

or am i stupid

[sleek lake]

seems true

[honest creek]

but (0,5) is not (0,0) right

[sleek lake]

0,5 is 5

[honest creek]

?

0 * 5 =0

[proud basin]

the x intercept is at 0 so that means (0,0) aka the origin

[sleek lake]

0,5 is x intercept

equals 5

[crystal hill]

If one of your intercept is at origin the other one is also at origin

[honest creek]

wait i’m dumb

[burnt bridge]

I'd like to see this magic line 😎

[honest creek]

i think i’m sleep deprived lol

sorry and thanks

my brain cells are gone

[burnt bridge]

lol get some rest

[honest creek]

too much sleep debt

😭

[burnt bridge]

Beautiful

[honest creek]

thanks

.close

[marsh citrus]

Channel closed

Closed by @honest creek

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[warm fractal]

x + y = sqRoot(14)
x - y = sqRoot(7)

Prove that x and y are the same values and determine the values.
Help

[main idol]

x and y can't be the same value of their difference isn't zero

[marsh citrus]

@warm fractal Has your question been resolved?

[warm fractal]

so this question is imposssible?

hold up

[sacred idol]

maybe send a picture of it?

[main idol]

As you've stated yes

[warm fractal]

i just writtwn one

what if

x and y equals to 1.87

[main idol]

,calc 1.87-1.87

[elfin berry]

Result:

0

[main idol]

Doesn't equal sqrt(7)

,calc sqrt(7)

[elfin berry]

Result:

2.6457513110646

[warm fractal]

thats weird

oh

so basically my marks got cut for an impossible question

.close

[marsh citrus]

Channel closed

Closed by @warm fractal

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crude blade]

Hi, In your opinion, is this enough proof?

[crude blade]

A tautology is a proposition that is always true. Let p, q and r be three
propositions. Using the laws of propositions or the truth table, show that
((p ∨q) ∧(r →s)) ⇔((p ∧r) →s) ∧((q ∧r) →s) is a tautology.

[marsh citrus]

@crude blade Has your question been resolved?

[crude blade]

<@&286206848099549185>

[cedar stone]

your truth table is wrong

[crude blade]

hm.. I felt confident with the table.
Which columns include mistakes?

[cedar stone]

p and r has mistake for certa

[crude blade]

I can't find the mistake in it. p and r to be true it has to be both true, right? there are only 4 instances when both are true.

What am I missing?

[cedar stone]

sry

i meant smthing else

F<=>F is T

[marsh citrus]

@crude blade Has your question been resolved?

[brave knoll]

Sir, first F<=>F is indeed true, hence a line must be corrected

[brave knoll]

Secondly you know it is a pain in the ### to rigorously check those tables, thus I invite you to identify what lines are wrong using this useful website checker :
https://truth-table.com/#((p | q)∧(r→s))%E2%86%94((p%E2%88%A7r)%E2%86%92s)%E2%88%A7((q%E2%88%A7r)%E2%86%92s)

When you identified what lines are wrong and if u still struggle to see your error, you can ping us back

Plus, I invite you (for your exam) to first check if one variable vector can do all wrong to your expression so we can easily prove it is not a tautology without doing to whole 2^#nb-variables lines tables

In here, FFFF leads to F, hence the expression is not a tautology

Also if you are allowed, one would recommend you to develop / refactor the expression as you wish to simplify your life (eg M->S is the same as not(M) or S ; a way to remember is if a robber tells "No one Move or I Shot" you directly understand if you Move, you get Shot)

[crude blade]

Thank you so much for the advices! I will start over with these tools and informations.

[marsh citrus]

@crude blade Has your question been resolved?

[crude blade]

I could isolate 3 wrong lines with the help of that website.
In 2 of them after rechecking every step I found a few mistakes in the second half.
The 3 wrong lines shared the same issue in the last step that I remembered wrong and thought only T<-> T is true.

Thank you for pointing out that something was wrong and the provided tools!

I will try to practice the simplification as well.

[crude blade]

How detailed should be my reasoning why is it not a tautology?

[brave knoll]

Depends on your teacher
"Take (p,q,r,s) = (F,F,F,F), we get expr1 = F, expr2 = T so finalexpr=expr1 operator expr2=F hence finalexpr is not a tautology"
I let you substitute with the right values/expression

We can always add more details you know
One can write the definition of a tautology
Then writes its negation to exhibit why we do that (yielding a counterexample) but it is pretty obvious

[crude blade]

Thank you for everything! 🙂

.close

[marsh citrus]

Channel closed

Closed by @crude blade

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[red nimbus]

Can someone check my work please?

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I just realied S_n is not cyclic for n >= 3

An element g of order 12! would mean that is a generator of S_12 but that would mean S_12 is cyclic which is contradictory

[glacial hedge]

what does corrolary 4.1 say?

[hollow glen]

there's a different way to see why the first is true

think of a way to construct such an element

[red nimbus]

I wrote what it said

but let me think

[hollow glen]

it just say |U| divides |G|?

[red nimbus]

For the last one I wrote a nice contradiction

[glacial hedge]

so it's Lagrange's theorem

which doesn't state the existence of any element

[deep anvil]

In context with his previous reply, he is using the structure theorem for cyclic groups.

Which as he noticed afterwards does not work since S12 is not cyclic.

[red nimbus]

There is another one Corollary 4.2.
Is G a finite group then there is for every g in G the order of g a divisor of |G| is more suitable

[glacial hedge]

yeah still useless

[red nimbus]

why

[hollow glen]

for every g, but not there exists a g

[glacial hedge]

does it state the existence of an element?
no, it gives a condition to it

[red nimbus]

By that there must exist such g

ah ok

[glacial hedge]

please highlight the part of 4.2 that asserts the existence of an element

[red nimbus]

doesnt

[glacial hedge]

right

[red nimbus]

it's basically stating what i need to show here

[glacial hedge]

no it's not

[red nimbus]

huh

[hollow glen]

maybe just forget about the theorem for now

[glacial hedge]

also, just don't use theorems
this one is best done by elementary means

[hollow glen]

it would only help to disprove there exists an element of order 13 or something

[glacial hedge]

as I said

[red nimbus]

How would I find such g? Clearly there are 12! possible elements

I wanted to apply theorem just to ensure the existence

[hollow glen]

go through all 479001600 of them

[glacial hedge]

are you just being introduced to Sn or do you happen to already know a few things about it, so that you understand what Sn is?

[red nimbus]

introduced, i am studying and reviewing

for the exam

S_n is a symmetric group which is about permutations afaik

[hollow glen]

can you give an element of order 3?

[red nimbus]

i see

[red nimbus]

<1,2,3>?

[hollow glen]

the cycle that rotates the first 3 elements? yes

[red nimbus]

I thought of it like [<1,2,3>] I would have <1,2,3>^n where it is equal to id if n = 3

So basically

g could be

3 * 5 * 4 then <1,2,3> o <1,2,3,4,5> o <1,2,3,4>

?

[hollow glen]

not like that

[red nimbus]

hmm ok

[hollow glen]

but almost

[red nimbus]

Do the orders need to be prime

instead

3 * 5 * 2 * 2

<1,2,3> o <1,2,3,4,5> o <1,2> o <2,3>

[hollow glen]

no but you overlapped them

[red nimbus]

what does this mean

[hollow glen]

the cycles are overlapping, no guarantee that their orders will multiply

[red nimbus]

oh

[red nimbus]

So you are saying I need disjoint cycles?

[hollow glen]

yeah

[red nimbus]

<1,2,3> o <4,5,6,7,8> o <9,10,11,12>

[hollow glen]

right, there is one more condition though that needs to be talked about

would <1, 2> ° <3, 4, 5, 6> be of order 8

[red nimbus]

would it be order 2?

[hollow glen]

just order 4

4 is their least common multiple

[red nimbus]

<1, 2> ° <3, 4, 5, 6>
but like
3 to 4 and 4 to 4 ... 5 to 6 and 6 to 6 would make id so I thought only <1,2> is left

[spark otter]

and 6 to 3

[red nimbus]

6 to 3 and 3 to 4 so <6,4>

<1,2> o <4,6>

[spark otter]

huh

what did you do

[hollow glen]

123456
214563
125634
216345
123456

[red nimbus]

and 3 to 3 so id again?

[spark otter]

and 3 to 4 you mean

I don't understand what you're doing

[red nimbus]

that was wrong

[spark otter]

are you considering this permutation ; <1, 2> ° <3, 4, 5, 6> ?

or the square of it

[red nimbus]

yes

I was considering <1, 2> ° <3, 4, 5, 6>

[spark otter]

well in this permutation

1 goes to 2, 2 goes to 1

3 -> 4 -> 5 -> 6 -> 3

one arrow at a time

[red nimbus]

<1, 2> ° <3, 4, 5, 6>
3 -> 4 and 4 -> 4
4 -> 5 and 5 -> 5
...
6 -> 3 and 3 -> 3
so
<1, 2> ° <3, 4, 5, 6> = <3> o <4> o <5> o <6> ?

[spark otter]

???

[red nimbus]

This is how I learnt it

[spark otter]

ok

so

[red nimbus]

you start from the right one at a time

and then look where it's mapped left

[spark otter]

why <3> <4>... though?

if 3 -> 4 and 4->4, then in total, 3 -> 4

if 4 -> 5 and 5->5, then in total 4 -> 5

[red nimbus]

wait i am dumb

<1, 2> ° <3, 4, 5, 6> = <3, 4, 5, 6>?

[spark otter]

still not

1 -> 1 and 1 -> 2, so 1 -> 2 in total

2 -> 2 and 2 -> 1, so 2->1 in total

so 1 and 2 move as well

[red nimbus]

<1, 2> ° <3, 4, 5, 6> = <3, 4, 5, 6> o <1,2> instead?

[spark otter]

well yeah they're the same thing

[red nimbus]

so id

[spark otter]

you just switched the order of disjoint cycles

which doesn't do anything

[spark otter]

?

[red nimbus]

e.g. <1,2> o <1,2> = <1> o <2> and I learnt that's equal to id

[spark otter]

well yeah that's true

1 -> 2 and 2 -> 1 so in total 1 -> 1

and 2 -> 1 and 1 -> 2 so in total 2 -> 2

this is equal to id

[red nimbus]

ok i feel like i missed out the point

[spark otter]

in this permutation however, when you account for all movements

in the end

numbers change place

[red nimbus]

what is the conditionother than disjoint cycles

[spark otter]

condition for what?

[red nimbus]

.

[spark otter]

it's about the order of a permutation, written in disjoint cycles

<1, 2> and <3,4,5,6> are disjoint

so the order of <1, 2> ° <3,4,5,6>

[spark otter]

is the lcm of their orders

because

(<1, 2> ° <3,4,5,6>)^k = <1, 2>^k ° <3,4,5,6>^k

[red nimbus]

I need to see if this is somewhere in the script

[red nimbus]

oh

Why is the order of a product of disjoint cycles equal to the least common multiple

is there a proof or is this just reasoning

[spark otter]

for <1, 2>^k ° <3,4,5,6>^k = Id

you need both <1, 2>^k = Id

[red nimbus]

OH

[spark otter]

and <3,4,5,6>^k = Id

[red nimbus]

K=4

[spark otter]

so k must be a multiple of both their orders

[red nimbus]

thanks

so i need to make sure lcd = 60

what if i just took

<1,2,3,4,...,60>

[copper raven]

if only we had 60 elements to permute...

we're in S_12 remember

you can't go up to 60 like that

[red nimbus]

lmao i am so done

[copper raven]

you already have a good example here

3 disjoint cycles

orders 3 4 and 5

what's the lcd of 3 4 and 5 ?

[red nimbus]

3 * 4 * 5 I guess

[copper raven]

yeah they're coprime

[copper raven]

i.e. 60

[spark otter]

the gcm is better

[red nimbus]

so writing 60 in prime factors (fundamental theorem of arithmetic) and ensuring to pick disjoint cycles is the take away i have, to find g here

that will be a very interesting exam, i feel like i am wobbling

anyway for the last one i did this

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

lcd(a,b,c) = |abc|/gcd(a,b,c) = 60/1 = 60 basically

thanks anyway

.close

[marsh citrus]

Channel closed

Closed by @red nimbus

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[red nimbus]

Let $G$ be a group and $a,b \in G.\$
Show [ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
It is $[a] = [b] \Leftrightarrow \forall b \in [b] : b \in [a] \text{ and } \forall a \in [a] : a \in [b].\$
Equivalently, there exists therefore $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.$

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

Can someone check my proof?

[devout mauve]

\forall b in [b]

[red nimbus]

huh

[copper raven]

lcm but yeah

[devout mauve]

so this needs fixing and then the rest will also be adapted

[red nimbus]

where

[devout mauve]

you cant write forall b when b is already introduced

[red nimbus]

OH

but

why did my professor say a = b^j and b = a^k

wait

nvm

[devout mauve]

a statement of the form "forall (variable) in (set that depends on that variable)" should immediately make you worry

[red nimbus]

i mean technically b is in [b] but I meant something else right

[odd crest]

for all c in [b], c in [a]

is better

[red nimbus]

ya

rein technically

[odd crest]

i actually wonder if you need both conditions

[red nimbus]

wouldn't it suffice to argue that b is in [a] already and a in [b] because of the equality

[devout mauve]

for one of the directions, yes

[red nimbus]

for one yes

[odd crest]

i don't know if there exists some equivalence classes [a] and [b] such that for all c in [a], c in [b] but not for all d in [b], d in [a]

[devout mauve]

the sets are the same, so they have the same elements

(but those arent equiv classes here)

[red nimbus]

yeyes

yes

Sorry Katharine it's another notation

[odd crest]

oh ok

[red nimbus]

Let $G$ be a group and $a,b \in G.\$
Show [ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
It is $[a] = [b] \Leftrightarrow \forall x \in [b] : x \in [a] \text{ and } \forall y \in [a] : y \in [b].\$
Equivalently, there exists therefore $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.$

[elfin berry]

𝔸dωn𝓲²s

[devout mauve]

why does that show both directions

[red nimbus]

I think I will redo the proof for both directions

cause now i am not sure anymore

[devout mauve]

a nice standard phrasing is "in particular for the choice x=b then ..."

warum machst du heini eigentlich am 24ten mathe

[red nimbus]

ich bin alleine und ich will ungern durchfallen sonst bin ich gezwungen mein studium hinzuschmeißen

[devout mauve]

gönn dir mal ne pause

[red nimbus]

ja gleich

[odd crest]

what notation is it

[devout mauve]

subgroup generated by the element

[red nimbus]

i dont get that

[devout mauve]

thats the argument you wanted to do here

for one direction anyway

I just meant that this phrasing is a standard way to do this argument

[odd crest]

in order to prove a biconditional you should prove both implications

so in your proof you should have a bit that proves =>

and a bit that proves <=

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I know, I just hoped I could do it without

I will never manage this in an exam

[regal coral]

in the first part, you can't choose x=b and y=a since you're trying to show something for all x and all y

[odd crest]

well not necessarily

[regal coral]

wait

i'm wrong

sorry

that wasn't what you're trying to show

[glacial hedge]

in <= the powers don't add, they multiply

[odd crest]

you should also say Suppose there exist

j, k in Z

such that

[glacial hedge]

yeah wording is bad

[odd crest]

or i like it better when you say Assume there exists

because you want to prove the implication

meaning you want to show that assuming P is true

then it results in Q being true

which proves P => Q

[elfin berry]

𝔸dωn𝓲²s

[odd crest]

$x = a^n = (b^j)^n$

[elfin berry]

Katharine

[red nimbus]

yeee

Do you want me to fight you??

😛 Fight me

[odd crest]

also i think you need to switch your a and b on the last line

because you define x as being generated by a

and then use the assumption

[glacial hedge]

no, you're not worthy of that much attention

[odd crest]

to show it is also generated by b

[red nimbus]

you are right!

[odd crest]

Also i'm guessing the definition of [a] = [b] is the bit after the implication

[red nimbus]

but technically nj = new whole number

[red nimbus]

yk nvm i

cant articulate my thoughts rn

[red nimbus]

Let $G$ be a group and $g \in G.$
[ [g] := {g^n \in G \mid n \in \mathbb{Z} } ]

[elfin berry]

𝔸dωn𝓲²s

[odd crest]

actually a bit better way of writing the consider bit is to say "Let x in [a]. Thus there exists some n such that x = a^n. Using the assumption gives x = b^nj. Hence for all x in [a], x in [b]"

something like that i think is more general or gives a more general vibe

as choosing n and m feels somehow like choosing a specific thing

[regal coral]

yeah i think that's more correct

[red nimbus]

wdym more corect 😭

I will come back in 10min

[regal coral]

to prove [a] = [b], show [a] is a subset of [b] and vice versa
to show [a] is a subset of [b], show for all x in [a], x is in [b]

so you start with "let x in [a]" rather than considering n in Z and defining x:= a^n

[marsh citrus]

@red nimbus Has your question been resolved?

[red nimbus]

ok i am bacc

[red nimbus]

ok i understand just a nicer way to say stuff

Let $G$ be a group and $a,b \in G.\$
Show [ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
$\Rightarrow :\$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].\$
Choose $x = b$ and $y = a.\$
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.\$
$\ \Leftarrow :\$
Let $x \in [a]$ and $y \in [b].\$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}.\$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

Is this now ok

I think it's ok

If I were the reader I would understand it

[odd crest]

it's difficult to use yourself as the benchmark

as you know everything about the proof

you know the intent behind the words

i would change the second to last line a bit

[red nimbus]

yea i know, but i can't expect people to be there all the time so i used my imagination

[odd crest]

it feels a bit awkward

[red nimbus]

i mean in an exam nobody will verify the proof too besides me

[odd crest]

but apart from it feeling a bit awkward i see the proof and accept the proof

actually

you should add the assumption back

[red nimbus]

what part is making you awkward

[odd crest]

very important

The line "Then for some..."

[red nimbus]

so it's not that obvious yet

[odd crest]

It reads awkward

[red nimbus]

hmm okay I thought it's short and concise

[odd crest]

Let $G$ be a group and $a,b \in G.$
Show [ [a] = [b] \Leftrightarrow \exists j, k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}$
$\Rightarrow :$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].$
Choose $x = b$ and $y = a.$
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.$
$\ \Leftarrow :$
Let $x \in [a]$ and $y \in [b].$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}.$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$

[elfin berry]

Katharine

Let $G$ be a group and $a,b \in G.\$
Show [ [a] = [b] \Leftrightarrow \exists j, k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
$\Rightarrow :\$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].\$
Choose $x = b$ and $y = a.\$ 
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.\$
$\ \Leftarrow :\$
Let $x \in [a]$ and $y \in [b].\$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}.\$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.53 $[a] = [b]$ implies $\forall
                                  x \in [a] : x \in [b] \text{ and } \forall...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```

[red nimbus]

wait

Show \[ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. \]
$\textbf{Proof.}\\$
$\Rightarrow :\\$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].\\$
Choose $x = b$ and $y = a.\\$ 
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.\\$
$\\ \Leftarrow :\\$
Let $x \in [a]$ and $y \in [b].\\$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}.\\$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$```

damn

I thought using the assumption is enough in a proof

not announcing that i will use it too

because a proof without the assumption is worthless?

[odd crest]

Let $G$ be a group and $a,b \in G.$
Show [ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
$\Rightarrow :\$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].\$
Choose $x = b$ and $y = a.\$
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.\$
$ \Leftarrow :\$
Assume that there exist some $j, k \in \mathbb{Z} \text{ s.t. } a = b^j$ and $b = a^k. \$
Let $x \in [a]$ and $y \in [b].\$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}. \$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$

[elfin berry]

Katharine
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[red nimbus]

here is the code btw

[odd crest]

Let $G$ be a group and $a,b \in G.\$
Show [ [a] = [b] \Leftrightarrow \exists j,k \in \mathbb{Z} \text{ s.t. } a = b^j \text{ and } b = a^k. ]
$\textbf{Proof.}\$
$\Rightarrow :\$
$[a] = [b]$ implies $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a].\$
Choose $x = b$ and $y = a.\$
Then there exists $j,k \in \mathbb{Z}$ s.t. $a = b^j$ and $b = a^k.\$
$\ \Leftarrow :\$
Assume that there exist some $j, k \in \mathbb{Z} \text{ s.t. } a = b^j$ and $b = a^k. \$
Let $x \in [a]$ and $y \in [b].\$
Then for some $n,m \in \mathbb{Z}$ it is $x = a^n = b^{nj}$ and $y = b^m = a^{mk}.\$
Therefore $\forall x \in [a] : x \in [b] \text{ and } \forall y \in [b] : y \in [a]$ which implies $[a] = [b].$

[elfin berry]

Katharine

[odd crest]

so that line is vital

because it's the assumption for proving the implication

[red nimbus]

almost read trivial

yea i know

i just thought when proving <= you already assume the assumption

[odd crest]

you could do it like that but for someone random

[red nimbus]

so i thought i didnt need to state it explicitly

[odd crest]

they will be confused

[red nimbus]

okokoko

[odd crest]

you know that when you write <= that you mean assuming Q

but someone else will just read the "Let .."

[red nimbus]

P <= Q
prove P using/assuming Q?

[odd crest]

yes

[red nimbus]

ok I got you

[odd crest]

the second to last line

the line

"Then for .."

i personally feel reading that line is a bit awkward

it's not fatal

it's understandable

and the proof can be understand without changing it

just a personal opinion

[red nimbus]

What would you change it to instead

[odd crest]

something like

Thus there exist some $n, m \in \mathbb{Z}$ such that $x = a^n$ and $y = b^m$.

[elfin berry]

Katharine

[odd crest]

And then adding a line where you say

something along the lines of

not exactly

because it's kinda cumbersum

Using the assumption this means $x = a^n = b^{nj}$ and $y = b^m = a^{mk}$.

[elfin berry]

Katharine

[odd crest]

what i wrote is cumbersome as hell

[red nimbus]

I prefer to keep it short

[odd crest]

you could probably get away with just adding the $= b^{nj}$ and $= a^{mk}$

[elfin berry]

Katharine

[red nimbus]

like it's my last proof I write on Earth

[odd crest]

To "Thus .."

so it becomes

Thus there exist some $n, m \in \mathbb{Z}$ such that $x = a^n = b^{nj}$ and $y = b^m = a^{mk}$.

[elfin berry]

Katharine

[odd crest]

but again it's a personal preference

[red nimbus]

and I respect it

thank you so much

.solved

[marsh citrus]

Channel closed

Closed by @red nimbus

Use .reopen if this was a mistake.

[marsh citrus]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[red nimbus]

hey gang

[red nimbus]

can someone please check out my work 🙏🏻

[elfin berry]

𝔸dωn𝓲²s

[regal coral]

isn't this just {e}?

[marsh citrus]

@red nimbus Has your question been resolved?

[regal coral]

because <a> is a subgroup so ord(<a>) divides ord(G) so ord(a) divides ord(G) so ord(G) = ord(a) * j so a^n=a^(k * j * ord(a)) = e^(k * j) = e

[odd crest]

What if G is not a finite group?

oh wait then the order is 0

[red nimbus]

0?

[regal coral]

the order is infinite i guess

[red nimbus]

yea

[odd crest]

nvm

then i'm not sure that your n_1 is a valid choice

also i think they ask for an example group

[red nimbus]

yea i forgot maybe I can choose Z_m

[odd crest]

the kernel of $\Phi_n$ is all $x \in G$ such that $x^n = e$

[elfin berry]

Katharine

[red nimbus]

yea

so i thought n = ord(G)k

now i understand

why u asked for finiteness

[regal coral]

G can be infinite i think

[odd crest]

$\ker{\phi_n} = \left{ x \in G \mid x^n = e \right}$

[red nimbus]

\{

do another slash

before {

[odd crest]

oh right

[red nimbus]

and \mid instead |

[regal coral]

ye you should say ker(phi) rather than ker(G)

[odd crest]

you need to escape }

[red nimbus]

anyway yea here e = 1 because we have multiplication i thought

[odd crest]

sorry

i'm dumb

[elfin berry]

Katharine

[regal coral]

nothing says you're working with numbers, G could be permutations or something

so use e i guess

[odd crest]

i like e more

personally

but you could call it 1

[red nimbus]

right

now i am dumb

[odd crest]

you haven't specified G so it could be something like S^n

which is permutations

but my book gives the identity of it to be 1

[red nimbus]

Does the choosing of G even matter for the kernel

[odd crest]

and all other elements to be some (a b c ...)

[deep anvil]

For abelian groups under multiplication it is customary to write 1

[odd crest]

anyway

[deep anvil]

And yeah the phi_n1 is the trivial function

But you have the right idea of trying to get the identity function

[odd crest]

just set n_1 = 0

oh no that doesn't necessarily work

[deep anvil]

That's also the trivial function

But you are close

[red nimbus]

n_1 > 1

[odd crest]

but you have to choose a group right?

[red nimbus]

So I thought choose n_1 = |G|

[red nimbus]

yea

non-trivial

I thought of Z_m

[odd crest]

Z_n

do you have to do multiplication groups?

[red nimbus]

no

just abelian

[deep anvil]

You proved it's the trivial function, not the identity function.

[red nimbus]

what's a trivial function

[deep anvil]

Mapping everything to 1=e

[odd crest]

Trivial function is

[red nimbus]

ah

[odd crest]

f: G -> G, a -> e

[red nimbus]

i got it all wrong again

[odd crest]

identity is f: G -> G, a -> a

[regal coral]

what about n_1 = 1

[odd crest]

not allowed

[red nimbus]

n_1 > 1

and n_2

[regal coral]

oh yeah

[odd crest]

maybe pick a dihedral group

[red nimbus]

n_1 = |G|+1 instead

[deep anvil]

Not abelian

[red nimbus]

ez

[odd crest]

ah fuck you're right

[red nimbus]

Ok let's choose Z_5

[regal coral]

yeah that should work

[deep anvil]

lmao that's kind of cheating but yeah

[red nimbus]

what am i supposed to do else sqrt(o²)

[odd crest]

Z_5 with n_1 = 3 works

but idk if any n_2 exists

[regal coral]

5 is prime

[deep anvil]

|G|

[regal coral]

so won't they all be isomorphic?

[red nimbus]

i chose prime purposely fuck

[odd crest]

n_2 = 5

then you got your answers

[regal coral]

oh yeah

[odd crest]

G = Z_5 with n_1 in {2, 3, 4} and n_2 = 5

i'm not entirely sure but i know 3 works

but i think 2 and 4 do too

[red nimbus]

Is 3 a generator of Z_5

[deep anvil]

Also nitpicky, but you should mention that (ab)^n=a^nb^n because of commutativity.

[odd crest]

yes

yes to kiyashi

ya?

not ya

[deep anvil]

yo

[regal coral]

3 should generate Z_5

[odd crest]

i noticed just as i pressed enter lol

[red nimbus]

so picking the generator makes it isomorphic

[odd crest]

anything but 0 generates Z_5 no?

[regal coral]

yeah

[red nimbus]

i didnt know this was commutativity but a^nb^n = b^na^n

[regal coral]

you have to use commutativity n times

[red nimbus]

bruh

[regal coral]

or something like that

[red nimbus]

this is now a line i wont draw ⚔️

[regal coral]

abababab = aaaabbbb

[red nimbus]

hell nah

[regal coral]

switch them all around

[odd crest]

you don't have to write it out

lol

[regal coral]

no but i'm just showing why it's commutativity

[red nimbus]

So actually I am mega dumb

I should have first picked a cool abelian group

and then proceed from there

[deep anvil]

yes

[odd crest]

This is fine apart from what kiyoshi said about saying that the reason it works is abelian groupness

this step only hold for abelian groups

[red nimbus]

now i understand...

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

I hope this is fine now

[regal coral]

i still think the kernel is wrong

the kernel is not always {e}

[red nimbus]

let me understand what you said before

[red nimbus]

hmm

because <a> is a subgroup so ord(<a>) divides ord(G) so ord(a) divides ord(G) so ord(G) = ord(a) * j so a^n=a^(k * j * ord(a)) = e^(k * j) = e

shit running out of time

[ \ker(\Phi_n) = { a \in G \mid \Phi_n(a) = a^0 = e } = { a \in G \mid a^n = e } ]

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

the same basically

[regal coral]

it's not the same

[red nimbus]

why

phi_n(a) = a^n by definition

[regal coral]

the set you described first was {a^|G|, a^(2|G|), a^(3|G|)...}

which is {e, e, e, ...}

[red nimbus]

I didnt write that

now

[willow owl]

if n=(k ord(G)) then a^n=a^(k ord(G))=[a^(ord(G))]^k=e^k=e regardless of what a and k are

[red nimbus]

wait

the kernel depends

if phi is bijective then we have the trivial kernel because phi is injective too

oh my

[regal coral]

yes

[red nimbus]

and everybody was quiet about that 😭

[regal coral]

ok i think

the kernel is all a whose order divides n

[red nimbus]

If phi is not injective then the kernel is not trivial 🗣️

[red nimbus]

[deep anvil]

there's also the problem you haven't addressed what happens when G is not finite

[red nimbus]

it is because i run away from problems i dont have the solution for

[sudden barn]

Is the ans to 6÷2 (1+2)... equal to 9 or 1 ?

Huhh

[red nimbus]

1 = 9 mod 8

[sudden barn]

Pls elaborate,am not the best at math

[cedar stone]

9

Just 9

[sudden barn]

But am getting 1 even my calculator is showing the ans as 1

[red nimbus]

your calculator interprets it as 6÷(2(1+2))

[sudden barn]

I also believe its 1 tbh...
Bare with me pls

[cedar stone]

Dyk bodmas

[red nimbus]

Since I helped you, can you also help me please

One hand washes another

[cedar stone]

Isn't the statement you are trying to prove wrong?

For example take group (addition modulo 3) and take n = 3

[willow owl]

isnt a^n=e for any a∈G with |a| | n?

i am trying to help but i have near 0 knowledge in group theory so

[red nimbus]

no worries ❤️

we are basically both on the same boat ❤️

[willow owl]

ohhh

[cedar stone]

Sry, I misread statement

[willow owl]

so doesnt this mean that n need not be a multiple of |G|

[regal coral]

yeah

[willow owl]

it just needs to be a multiple of |a|

[red nimbus]

yea but i dont get why thats the case

[willow owl]

so as long as a has finite order that satisfies this it should be in kerφ

[red nimbus]

how we get there

[willow owl]

you need a^n=e

the least exponent where a raised to it gives e is |a|

[red nimbus]

so not 0

ok

[willow owl]

the wording is not too good but do you agree with this ?

[red nimbus]

yes

[regal coral]

0^n = 0

[red nimbus]

i just accept continue

[regal coral]

the order of 0 is 1 and 1 divides n so 0 is in the kernel

[red nimbus]

ah

yea

[regal coral]

i mean assuming this is an additive group

[willow owl]

ok so if n=k|a| then a^n=a^(k|a|)=[a^(|a|)]^k=e

[red nimbus]

If $n = k\abs{a}$ then $a^n = a^{k\abs{a}} = ({a^{\abs{a}})}^k = e$

[elfin berry]

𝔸dωn𝓲²s

[red nimbus]

that was kinda easy

[willow owl]

in general, this provides elements in the kernel different than e

[red nimbus]

but like

a^|a| = e?

[willow owl]

yes

isnt that the defintion of |a|?

[regal coral]

by definition

[red nimbus]

I had this for example

which just contains e

{e}

but wouldn't n = k|a| also just contain e

[regal coral]

yeah

[cedar stone]

No?

[regal coral]

oh no

[cedar stone]

For example take addition mod 6

N = 2

[regal coral]

no it would

[willow owl]

you are not looking for a^n

[regal coral]

yeah

[willow owl]

you are looking for a∈G such that a^n=e

thats all

[red nimbus]

so

I actually left out possiblities

[willow owl]

$\ker\varphi={a\in G|\varphi(n)=e}$ so in our case $\ker\varphi={a\in G|a^n=e}$

[red nimbus]

yea

[regal coral]

yeah you shouldn't have a^n on the left of the bar |

[red nimbus]

.

[willow owl]

exactly

then to give a characterization of these a

you used the fact that |a| is the least number n such that a^n=e

then it follows that if n=k|a| then a^n=e and with this you have identified what are the elements a that are in kerφ

by saying that kerφ consists of all a∈G such that |a| | n ie such that n=k|a|

[regal coral]

and you should also say
if |a| doesn't divide n, then |a|k+r = n for some 0<r<|a|, so
a^n = a^(|a|k + r) = a^(|a|k) a^r = e*a^r = a^r, but since 0 < r < |a|, a^r is not e and therefore a^n is not e and so a is not in the kernel

[red nimbus]

https://tenor.com/view/am-i-cookedd-am-i-cooked-gif-7478847585523279700

[deep anvil]

Another tiny thing is that in the examples you give, you are using the additive structure of Z_5, but are writing the operations multiplicatively which might lead to confussion. For example, instead of writing a^5, you should be writing 5a.

[red nimbus]

i know

id id al wro

[willow owl]

btw if a∈G then a^k∈kerφ for any k∈N

[red nimbus]

what how

[regal coral]

hmm i don't think so

[willow owl]

and all of this works for any group G whether it is finite or infinite