User Ken Ono | chat.stackexchange.com

Ken Ono

Oct 17, 2018 17:11

I mean the formula $|G|=[G:H]\times |H|$

Ken Ono

Oct 17, 2018 17:09

@MikeMiller As, $\mathbb{Q}$ is not finite, counting formula will not work here, right?

Ken Ono

Oct 17, 2018 17:04

@MikeMiller I am still not getting how this lemma will help to answer my question :(

Ken Ono

Oct 17, 2018 16:52

@MikeMiller Got it :) thanks

Ken Ono

Oct 17, 2018 16:50

@MikeMiller So, I need to consider the equation $e^{2\pi i t}=1$? and by De Moivers, it is true for $t\in \mathbb{Z}$?

Ken Ono

Oct 17, 2018 16:48

Consider the action of $\mathbb{R}$ on the unit circle $S^1=\{z\in \mathbb{C}~|~~|z|=1\}$ by $t\cdot z=e^{2\pi it}\cdot z$. I want to find the $Stab(1)$, $Stab(1)=\{t~|~~e^{2\pi i t}\cdot 1=1\}=\{t~|~~(e^{2\pi i})^t=1^t=1\}=\mathbb{R}$. I am actually not sure about the last equality. Am I correct that $1^{irrational}=1$?

Ken Ono

Oct 17, 2018 15:57

@MatheinBoulomenos Thanks :-)

Ken Ono

Oct 17, 2018 15:53

Is it true that: "$S_n$ contains all groups of order $\le n$"? As, $S_n$ contains the group of order $n$ by Cayley theorem and as $S_n$ contains all elements of $S_{n-1}, S_{n-2},S_{n-3},\cdots, S_2$, I think the answer is true. Can anyone tell me I am correct or wrong?

Ken Ono

Oct 16, 2018 03:11

@BuddhiniAngelika or here - groupprops.subwiki.org/wiki/GAP:DirectProduct

Ken Ono

Oct 16, 2018 03:10

@BuddhiniAngelika Just search online first. gap-system.org/Manuals/doc/ref/chap49_mj.html

Ken Ono

Oct 16, 2018 03:09

@MikeMiller Ok, I will try to prove it. Thank you :-)

Ken Ono

Oct 16, 2018 03:04

@MikeMiller (i) $e\ast s=s$, $s\in S$ and (ii) $(g_1\cdot g_2)\ast s=g_1\ast (g_2\ast s)$ , $g_1,g_2\in G$

Ken Ono

Oct 16, 2018 03:03

which satisfies two properties

Ken Ono

Oct 16, 2018 03:02

@MikeMiller $G$ is a group and $S$ be a set, we say "$G$ acts on $S$" if there is a function $G\times S\to S$

Ken Ono

Oct 16, 2018 03:00

@SharathZotis Intro to prbability by Sheldon Ross

Ken Ono

Oct 16, 2018 02:57

@MikeMiller I know the foraml definition of action as defined in books, can you explain abstractly, which will help to think in different way?

Ken Ono

Oct 16, 2018 02:56

@MikeMiller It's true :-) I am learning it first time.

Ken Ono

Oct 16, 2018 02:53

as the action is not given, not getting how to treat $q\ast a=a$, to get the stabilizer. But, as $\mathbb{Q}$ is group under operation $+$, I think the operation need to be $+$, otherwise it will not satisfy the property $(q_1\cdot q_2)\ast a=q_1\ast(q_2\ast a)$, where "$\cdot$" is the group operation.

Ken Ono

Oct 16, 2018 02:43

How can I show if there is any action of $\mathbb{Q}$ on the set $\{a,b,c,d,e\}$ such that $\text{stab}(a)\neq \mathbb{Q}$?

Ken Ono

Oct 13, 2018 15:36

@LeakyNun Thanks

Ken Ono

Oct 13, 2018 15:34

@LeakyNun I am confused little bit, do you mean, $O_{s_0}=S$ , so, for any $s\in S$ we have $g$ such that $s=g~s_0$ ?

Ken Ono

Oct 13, 2018 15:32

$\text{stab}(gs_0)=g~\text{stab}(s_0)g^{−1}$ - can u explain this step?

Ken Ono

Oct 13, 2018 15:27

It is told that G is abelian. It seems you are not using that fact

Ken Ono

Oct 13, 2018 15:25

@LeakyNun Where we using that the group is abelian?

Ken Ono

Oct 13, 2018 15:23

I am not getting any idea how to proceed

Ken Ono

Oct 13, 2018 15:22

Let $G$ be an abelian group acting transitively on a set $S$. Consider an element $s_0\in S$, if the $\text{stab}(s_0)=H$ then, can we say $\text{stab}(s)=H$ for any $s\in S$?

Ken Ono

Oct 13, 2018 15:18

@TobiasKildetoft Also, there is a 6 , and as order of orbit have to divide order of group, it's impossible

Ken Ono

Oct 13, 2018 15:16

@LeakyNun But, the group may act on a set(which don't have identity)

Ken Ono

Oct 13, 2018 15:14

@LeakyNun Why you need 1?

Ken Ono

Oct 13, 2018 15:13

@LeakyNun Oops! forgot that

Ken Ono

Oct 13, 2018 15:12

@LeakyNun Yeah I have got same examples for iv) and v)

Ken Ono

Oct 13, 2018 15:11

Which of these CANNOT be the class equations for a group of the appropriate order?
i)10 = 2 + 2 + 2 + 2 + 2
ii)15 = 1 + 3 + 5 + 6
iii)4 = 1 + 1 + 2
iv)6 = 1 + 1 + 1 + 1 + 1 + 1
v)6 = 1 + 2 + 3 I think for problem ii) we can't find such class equation, as $6$ can't divide $15$, so a group of order $15$ can't produce a orbit with cardinality $6$. Rest problems I think can be class equations. Can anyone check it?

Ken Ono

Sep 25, 2018 12:51

Yeah not abelian

Ken Ono

Sep 25, 2018 12:49

@TobiasKildetoft Thank you but why that helps?

Ken Ono

Sep 25, 2018 12:46

Let, $G=GL_n(\mathbb{R})$ , the group of all invertible $2\times 2$ matrices and $H=\{\begin{bmatrix} 1&0\\0&1 \end{bmatrix}, \begin{bmatrix} -1&0\\0&-1 \end{bmatrix}\}$. How can I show $G/H$(qutiont group) is abelian or not?

Ken Ono

Mathematics

$Mathematics$ Mathematics