/sci/ - Science & Math

[Anonymous No.2304469]

Hi sci
you are my last resort. Is there a rule of thumb for finding every nonequal element of a finite group with its definig relations. Like a^4=1; b^3=1; bab=a
Pic realted, i feel quite derp about it but i think i've just found over 30 nonequal elements with theese relations. That cant be right becouse i need to make a multiplying table with them and that woud be freaking huge, no teacher can be that sadistic.

[Anonymous No.2304478]

>>2304469
You should be able to logic this out.

[Anonymous No.2304480]

The only element here is 1.
No,seriously.

[Anonymous No.2304481]

>>2304480
From what i see without thinking about it, elements are 1 and sqrt4(a)

[Anonymous No.2304484]

>>2304480
How did you get to that?

[Anonymous No.2304486]

>>2304481
You are DEFINITELY not thinking about it.

[Anonymous No.2304489]

>>2304484
Well, 1^x=1 with any x.
So, a=1 and b=1 is possible, since bab=a still applies.
Didn't think about anything else really, just what came up my mind the fastest.

[Anonymous No.2304496]

>>2304489
I think a and b are not equal, else this coud be applyed to any definig relations.

[Anonymous No.2304504]

a^4=1; b^3=1;
So, a^4=b^3.
Since a=bab, we get
(bab)^4=b^3
(b^2*a)^4=b^3
b^8*a=b^3
a=b^3/b^8
a=1/b^4
And THAT'S what we have been looking for. for any number b!=0, a will be equal 1/b^4

[Anonymous No.2304511]

>>2304504
The problem with that is that a and b are not commutative ab!=ba, maybe i've forgot to mention that but commutativity isnt in the group axioms (I think thats what they called in english :))

[Anonymous No.2304514]

>>2304504
You forgot that a must be a>=0, since it is raised to an even exponent.

[Anonymous No.2304519]

Op again.
with bab=a we get baba^3=1 and a^3bab=1. So any string of a and b wich doesent contain baba^3; a^3bab a^4 and b^3 isa unique element, wich gives us infinite elements. Proove me wrong

[Anonymous No.2304520]

>>2304511
Damn, i don't understand what commutative means...
I'm Russian and don't know english very well

[Anonymous No.2304522]

>>2304520
It means ab is not equal to ba. I dont know either if that's the right word for it in english

[Anonymous No.2304531]

>>2304522
And how is THAT possible? Any two numbers multiplied result the same, no matter the order you write them in, that's an axiome. Something is wrong here, that can't be possible.

[Anonymous No.2304532]

>>2304531
>he thinks everything behaves like real numbers

Try multiplying two matrices, see how that works for you, bro.

[Anonymous No.2304535]

>>2304532
>matrices
>in a task that requires you to write a mutiplication table with it's results
Yeah, the teacher must be a gigantic douchebag then.

[Anonymous No.2304536]

I want someone who has some understanding of abstract algebra dammit :(

[Anonymous No.2304538]

>>2304469
I'm kinda sad that no one has given you any real help yet. Check out page 44 of "Algebra" by Artin. (very similar problem)

The last defenying relation implies you can write all combinations in the form (a^n)*(b^m). With n and m from 0 to 4 (and 3) (respectivly).

Write out all the combinations, then reduce duplicate results using your defnying relations.

Make sense?

[Anonymous No.2304539]

>>2304536
For simple abstract algebra, see this
>>2304504

Also, abstract algebra is probably the most entertaining part of maths, it is always fun to argue about something regarding it.

[Anonymous No.2304542]

>>2304531
>Thinks everything commutes

LMAO, how old r u?

[Anonymous No.2304544]

There are 12 elements:
1, a^3, a^2, b, a, a^3*b, a^2*b, a^3*b*a, a*b, a^2*b*a, a*b*a, b*a

[Anonymous No.2304545]

>>2304542
Ich bin 8 und was ist das?

[Anonymous No.2304548]

>>2304544
Is there some basic method you get that with?

[Anonymous No.2304552]

>>2304538
I think its not just (a^m)(b^n) but (a^m)(b^n)(a^o)(b^p)..... to the infinity with m,o,... under 4 and n,p,... under 3, and i have to simplify theese with the third relation.

[Anonymous No.2304555]

>>2304544
Nope,

(b)(a)= (a)(b^2)

All the combinations are usually written in
a^n * b^m form

[Anonymous No.2304559]

>>2304552
No, it is a^n * b^m, I gave you the reference already. You should be able to find it on the net.

What part are you not understanding? Why do you assume these huge numbers?

[Anonymous No.2304560]

>>2304548
The basic method in easy examples like OP's is to notice every elements is in the form <span class="math">a^u b^v a^w[/spoiler] with u,v<4 and w<3. List all 16 elements you get and remove those you can simplify with your relations. There's also a general algorithm to enumerate elements of a finitely presented group but it's not convenient to use by hand. You can use GAP (gap-system.org) to do the computations for you.

[Anonymous No.2304568]

>>2304559
You are being arrogant in your mistake. It doesn't matter how you name your elements, write ab² insteal of ba if that makes you happy but for enumerating them you can name them however you want.

[i am very upset No.2304573]

shouldnt there be 12 elements? (a^m goes 0..3, b^n goes 0..2 and every 'other' multiplication is reduced by the ba = ab^2 rule)
totally shredded my symmetry box

[Anonymous No.2304576]

>>2304573
Yes there are, exactly for the reason you state.

[Anonymous No.2304577]

>>2304568
>It doesn't matter how you name your elements

Well then why don't you just use a^-1 and b^-1 in there as well? LOL

You are correct, that there are an infinite amount of ways to write this shit. However, the standatd way to write that shit is with positive exponents, and to completely commutate out the operators. It will make further mathematics much easier.

[Anonymous No.2304583]

>>2304573
Glad you finally got it OP

[Anonymous No.2304584]

>>2304577
ba is shorter than ab², I should be the one laughing at you for using uselessly long names.

[Anonymous No.2304587]

>>2304584
Yeah, and a^-1 and b^-1 will be shorter then whatever other version you use. You still don't use them though.....LMAO.

Have you even taken group theory?

[sage No.2304589]

>>2304584

[Anonymous No.2304592]

>>2304584
WTF AM I READING?

[Anonymous No.2304594]

>>2304592
>>2304589
>>2304587
You had to samefag when you didn't even figure out the guy was mocking your stupidity in the first place

[sage No.2304600]

>>2304594

[Anonymous No.2305541]

OP, are you still here?

[Anonymous No.2305598]

the presentation you have it's from the finite group T http://mathworld.wolfram.com/FiniteGroupT.html, and as it has been said, it has 12 elements