First of all thanks for being here :D ill upload the pic now :)

I don't understand it well in order to solve it yet

Because did not read that far, I just wonder if it is solvable, ie , no mistakes on that question :)

am currently working on this now , I did part a) was easy

but I dont understand well how part b) differ from a)

I mean i could find an element of S_6 that makes p_1(a) and p_2(a) conjugate for all a in (Z/2Z)^2

(a) $\forall a \exists \sigma:~ \sigma\rho_1(a)\sigma^{-1}=\rho_2(a)$

(b) $\exists\sigma\forall a:~ \sigma\rho_1(a)\sigma^{-1}=\rho_2(a)$

aha

"For all people A, there exists a person Sigma such that Sigma gave birth to A" versus "There exists a person Sigma, such that for all people A, Sigma gave birth to A."

okay so they asking for a sigma that conjugate all of em :D

Ehm donno how to prove this but ill Think about it now

asking you to prove a sigma doesn't exist

yes I got that :)

think about how the corresponding images act on {1,2,3,4,5,6}. conjugate subgroups should act the same, i.e. have the same number of fixed points, orbits, etc.

But ehm, in part a) I did it by computing that sigma is that the only way ?

two permutations are conjugate iff they have the same cycle type

use that for (a)

So I dont need to write explicitly what sigma is?

depends on if by show they mean "prove" or "illustrate with examples"

I Think Ill just put what you said , and add that to what i have written

because the way I did it was, i figured out what p_1 , p_2 of each element is

e.g p_(1,0) = sigma p_2(1,0) sigma' = (13)(24) sigma (12)(56) sigma'

then mapped 1-->1 , 3-->2 2 -->5 ,4-->6

(235)(46) is example of such element in S_6

I Think putting the while calculation and that elements in S_n are conjugate if they are of same cycle type should be a well written prof :D

Can you please explain what corresponding images act on {1,2,...}mean?

corresponding images i dont understand well

the representation is a function from the group into S_6, so it has an image which is a subgroup of S_6, which acts by permutations on {1,2,3,4,5,6}

allways the way you say things is on higher level :D i really love that :D

thanks

Ehm anon if you are ok with this, can you go through the definitions with me ? we only did 2 lectures and I really want to see if I got them right

this is very new to me

i Think that way , id be able to go further alone, without asking alot of stuff =p because that is my main problem, not understand or worst missunderstand defintions

If we start with what is a representation ?

what I understood so far, it is a homomorphism between a Group G, and GL(V)

also it is linear

yes. one can abuse notation by calling V the representation

what is the correct way to write it?

G->GL(V) being a homomorphism is the same as G having a group action on the vector space, such that for all g in G, the map v->gv is a linear map

I guess $(V,\rho)$ where $\rho:G\to GL(V)$ is the homo would be the most formal way to write it. But few people say "let $(G,\bullet)$ be a group..." instead they just say "let $G$ be a group," and similarly people can just say "let $\rho$ be a rep" or "let $V$ be a rep" of $G$

and why would that call it "representation " ?

I want to see the big Picture of this subject, because I really enjoy algebra , much Clean stuff than analysis or other area

from what I understood now, we have a Group action on a set with a structure now , vector space, not like Before G acting on X ( set )

I think of the images $\rho(g)$ of the elements $g\in G$ as "ambassadors" to the nation of $V$ from the homeland of $G$

so, like representatives

aha neat :D

for each g in G, we can associate it with a linear map ( matrix ) GL(n,K)

I Think using GL(n,K) is better, n by n matrices over a field K,

better than GL(V)?

yes imo

what do you Think ?

I guess for certain calculational stuff, but usually I think of it as GL(V) not GL(n,K)

we are doing only finite dim vector spaces and also finite groups

mmhmm

its an intro course for us :D

@anon if you would give a big Picture of this subject in few Words what would it be ? :D

sometimes I think of a rep geometrically (as a generalization of symmetry groups of polyhedra, for instance), and sometimes I think of a rep algebraically (as a C[G]-module, where C[G] is the group algebra)

maybe wait till the group algebra for the second part

Oh right >< that is chapter 6 is Serres book

am trying to solve part b) now using that hint you gave me =p

anon if you have good tips for me in general please share some :D I really want to learn alot of algebra, very cool subject :D

I don't have tips in this generality.

haha okay :D

just to be clear, rho: Z/2Z x Z/2Z ---> S_6

since rho is a homomorpshim

Im(rho) is a subgroup of S_6

mmhmm

How does one figure out the image of rho

._.

we gonna use the idea of subgroup acting on the Group right

grrrrrrr

I see it but I dont see it ><

when you said it , i thought i had it

can you tell what the fixed points are automatically just by looking at the generators?

generators of what?

the subgroups in S_6

like, the first subgroup (image of rho_1) has (12)(34) and (13)(24) in it, and is generated by them

so what can you tell are fixed points of that subgroup?

5,6 no ?

i mean generated by those elements is V_4

right, 5 and 6 are fixed points

and 1,2,3,4 are not fixed points

image of rho_1 is : (12)(34) , (13)(24) , (14)(23) , (1)

what about the second subgroup (image of rho_2) now?

well there we dont have anything fixed

we have this : (12)(34) , (12) (56) , (34) (56) , (1)

as the image of rho_2

mmhmm, no fixed points

now, call these subgroups H and K

okay

if there is a permutation $\sigma$ such that $\sigma H\sigma^{-1}=K$, and $x$ is fixed by $H$, then $\sigma x$ would be fixed by $K$

note that if $\sigma\rho_1\sigma^{-1}=\rho_2$ as functions $G\to S_6$ then $\sigma\rho_1(G)\sigma^{-1}=\rho_2(G)$ as subgroups of $S_6$

am tyring to make sense of this please one moment

did not get it :(

are we alloed to use that Product that way ?

product?

Subgroups and elements of S_6

sigma H sigma'

Yeah, $\sigma H\sigma^{-1}=\{\sigma h\sigma^{-1}\mid h\in H\}$ is a subgroup conjugate to $H$

what you saying is, that if H fixes x , then

aha right

okay so your arguemnt is that, if such sigma would exist , that would mean that H anbd K are conjugate

but hmm

the second line I did not really get

note that if $\sigma\rho_1\sigma^{-1}=\rho_2$ as functions $G\to S_6$ then $\sigma\rho_1(G)\sigma^{-1}=\rho_2(G)$ as subgroups of $S_6$

If $\sigma\rho_1(a)\sigma^{-1}=\rho_2(a)$ for all $a\in G$, then $\sigma H\sigma^{-1}=K$ where $H=\rho_1(G)$ and $K=\rho_2(G)$

Since $\sigma H\sigma^{-1}=\sigma\{\rho_1(a)\mid a\in G\}\sigma^{-1}=\{\sigma\rho_1(a)\sigma^{-1}\mid a\in G\}=\{\rho_2(a)\mid a\in G\}=K$

aha i see what you did there =p

I understood that step that is one positive thing :D

but to argue that this cannot happen

I dont know how to put it in words

proof by contradiction

we're assuming there is such a sigma

so now we have $H$ and $K$ conjugate

which tells us if $H$ has a fixed point $x$, then $K$ would have a fixed point $\sigma x$

but $H$ has fixed points and $K$ doesn't, a contradiction

Very elegant Anon :D thanks alot ! :D

I need to learn how to aruge this way =p

I thought of saying that, H and K are hm

anyway, it is not good ><

Okay :) ill keep working on more exercices :D

You are the best anon :)

np