Groups

[old citrus]

I have to check if that's a group

[wind ocean]

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[old citrus]

But like is this the same as saying like

G = {((a+b)^2)/(a*b): a, b ∈ Q}

Because a different set looked like this and it was more understandable that way

[orchid whale]

if it were a group what would be it’s identity element ?

[old citrus]

What's an identity element

Like this in the other example?

[orchid whale]

a group is a set with a binary operation (here it’s o)

The operation o must be associative, have an identity e which satisfies e o g=g and g o e=g for any g, and each element has an inverse for this operation

ie for any g there is some h such that g o h=h o g=e

[orchid whale]

Here the binary operation is multiplication

It’s associative, the identity element is 1 and the inverse of 5^k is 5^-k

[old citrus]

I've done this one yes I just showed it cuz it looks much more understandable than the other one

[orchid whale]

So G is a group

[orchid whale]

Here your binary operation is a bit weirder so it may be confusing

[old citrus]

a and b are both positive rational numbers right

[orchid whale]

yes

[orchid whale]

Do you think G is a group for o ?

[old citrus]

So is it the same as saying something like this

[old citrus]

Well a o b is a positive rational number so the first condition is met

The 2nd oneee

Idrk how to check the 2nd condition

Like (x o y) o z = x o (y o z) but Idrk what to make these x y z equal to

[orchid whale]

If it had an identity element what could you say ?

call it e for example

you know that e o a=a for any a

what happens if you set a=…

Fill the dots

[old citrus]

Do you mean a=.. or e=..

[orchid whale]

e is fixed

a is a variable

e being an identity element you have e o a=a for any a

what happens when a=e ?

[old citrus]

Then it's ((e+e)^2)/(e*e) = e

[orchid whale]

Yes

you can calculate the left term

[old citrus]

So 4e^2 / e^2

[orchid whale]

Yes

[old citrus]

So 4 = e

[orchid whale]

so if e were an identity you would have e=4

can you see a contradiction ?

[old citrus]

But what's an identity element exactlyy

[old citrus]

Well you said for any a..

[old citrus]

Here

But we chose it to be e specifically

[orchid whale]

i assumed there was an identity element e

if G were a group by definition it wold exist

*would

Do you agree ?

[old citrus]

BUT LIKE WHAT'S AN IDENTITY ELEMENT LIKE WHAT DOES THAT MEANN

[old citrus]

What's the identity element here?

Like for example yk

[orchid whale]

e is an identity element if e o g=g o e=g for any g

that’s one of the properties of a group that you should have seen

It’s in the definition

[old citrus]

Oh so it's like

A neutral element?

[orchid whale]

Yes

[old citrus]

Okayy

[orchid whale]

Maybe my naming was what was confusing

Sorry

[old citrus]

Yes G would be a group if those 3 properties would be met so the existence of e yes by definition

[old citrus]

It's oke

[orchid whale]

Yes, so we’ve established that e=4 here

if G were a group with neutral element e then e=4

[old citrus]

Yes

[orchid whale]

now you can get a contradiction from here

Can you see how ?

[old citrus]

Well..

If e = 4 then it means 4 o g = g and 4 = g

[old citrus]

But you said any g so if we say it's anything else than 4 then won't it contradict this?

[old citrus]

For g = 3 this wouldn't be true

[orchid whale]

For any g

[old citrus]

Or just any other g than 4

[orchid whale]

Yes

You can see that 4 o 1 is not 1 for example

So what can you conclude ?

[old citrus]

It means it's not a groupp

[orchid whale]

Yes!

[old citrus]

But can we check the other 2 properties just for a bit practice :so

😭

[old citrus]

Other 2 as in for it to be an Abel group Idk what you call it in English

[orchid whale]

Abelian ?

[old citrus]

Yes thatt

[orchid whale]

Well it’s not a group

It being abelian means the binary operation is commutative

[old citrus]

So a o b = b o a it's that one right

[orchid whale]

Yes

Here it’s clearly the case

[old citrus]

Yes because addition and multiplication are both communicative

So wait what are the 3 conditions for it to be a group

First is x o y must belong to G

2nd is

x o (y o z) = (x o y) o z

3rd is x o x' = x' o x = x

[old citrus]

And this for it to be Abelian

[old citrus]

I FIND it confusing like what I should let all these numbers equal

[orchid whale]

it has an element x’ such that x o x’=x’ o x=x for any x ie it has a neutral element

[old citrus]

Especially with the case with x y z

[old citrus]

What is ie you said that before

[orchid whale]

Yes, here it does not hold so it can’t be a group

[orchid whale]

Abelian is specific to groups so here it’s better to say the binary operation is commutative

[old citrus]

Oh alrr

Can we try this one

a and b are integers.. so

Well a o b is an integer yes so it belongs to G

2nddddddd

IF we wanted to write something like (a o b) o c how would we write that??

[orchid whale]

Write a o b explicitly call it h

[old citrus]

We'd write (((-1)^a) * b + ((-1)^b) * a) o c

[orchid whale]

Then you calculate h o c

[old citrus]

Yes

Ohh okay so

Okay I see

[old citrus]

I'll do that in my notebook wait

Omg it's so long

[orchid whale]

Gl

Im too lazy to write it out

[old citrus]

Okay I don't think it means the 2nd condition

But okay okay

Thank youu then

Byebye

+close

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