last day (353 days later) » 

Jim
10:52 AM
@ThomasKlimpel , Hi.
I am studying group theory , I have some novice questions to ask, would you mind to join?
 
11:09 AM
Hi
 
Jim
Thanks for joining!
0
Q: Sylow Subgroup of a Symmetric Group acting on Non-Symmetric Matrix

JimI am studying algebraic graph theory with a shaky basic. As I am new to the topic, I would be thankful if anyone helps me to understand the following proposition. $A, B$ are matrices of size $m \times n$ (not symmetric matrices). Given, rows of $A, B$ are fixed, then each of them can have $...

 
I defer all this to Tobias and Thomas if they are interested. My algebra is not really very good.
 
Jim
Would you comment on above post?
@BalarkaSen It does not need to be rigorous!
 
I am not really interested, sorry.
 
@Jim I have a very hard time following what you are trying to ask there
 
11:13 AM
Why did I receive invitation to this room? And who did invite me?
 
Jim
@BalarkaSen it's ok , thanks for informing.
@MartinSleziak I did , is there any protocol to send invitation?
 
I know almost nothing about Sylow subgroups and finite groups. So I am not really suitable candidate for this discussion.
So it seemed to strange that you invited me - a user who has nothing with the topic of discussion. It seems rather random. (I do not know whether you have invited other users.)
 
Jim
@TobiasKildetoft i am sorry for you inconvenience , would you please tell me exactly which line/paragraph/statement is not well written?
 
@Jim Well, for one thing, you seem to start with $A = B$ but then talk about whether the columns will be arranged the same, which they would already be
 
BTW there used to be a chatroom for finite groups, but it is not active anymore.

 Finite Group Theory

Let this be a place to come together in a finite group and di...
 
Jim
11:19 AM
@MartinSleziak invitation was sent based on another chat room+profile , please feel free to leave.
 
@Jim That is a really generous offer. :-)
 
Jim
@TobiasKildetoft 'same' does not mean '=' , since matrices are not symmetric the arrangement /permutation of row and column are independent .
 
@Jim Then what do you mean by the matrices being the same?
The question is full of unexplained terms. Also, I am not sure what being symmetric has to do with this
 
Jim
$A$ is another permutation of $B$.
 
@Jim permutation in what sense?
 
Jim
11:25 AM
permutation of columns.
 
@Jim Then would that not already imply that they also satisfy the second part?
 
Jim
"2.and have same row arrangement/ permutation " tells that only columns are moving not rows.
 
But was that not what you just said "same" meant?
 
Jim
@TobiasKildetoft the relation ship between $A,B$ is $g * A = B$ where $g$ is a permutation matrix.
 
@Jim And $g$ permutes only columns?
 
Jim
11:31 AM
yes
 
And you seem to be asking whether there will always be some way to apply some element from $H$ such that at least one column becomes the same. Is that correct?
 
Jim
more than one column, yes.
 
do you mean at least one, or strictly more than one?
 
Jim
1. one is ...more like trivial + $H$ will act definitely more than 1 column. 2." there will always be some way to apply......." ->Actually , I am telling u the way, i would like to know whether it is valid or not
 
@Jim Please write complete sentences. It is very hard to understand what you mean otherwise.
 
Jim
11:37 AM
edited.
 
edited what?
 
Jim
1. $H$ will act definitely more than 1 column.
@TobiasKildetoft 2." there will always be some way to apply......." ->Actually , I am telling u the way, i would like to know whether it is valid or not
 
That is nowhere near a complete sentence
 
Jim
@TobiasKildetoft I mean more than one.
 
@Jim Hmm, then my guess would be that we need $p$ to not be too small, but let me think a bit
So we are really asking about the number of elements fixed by something of the form $hg$ for a fixed $g$ and with $h$ varying over $H$
 
Jim
11:45 AM
@TobiasKildetoft you can assume $p$ is not too small or consider the case where it is not too small.
 
Hmm, I think we can construct $g$ such that there is precisely one fixed point for all $h\in H$ for any $p$, with $n = 2p-1$. And it seems that if $n \leq 2p-2$ then there should be at least $2$ fixed points
Hmm, now I am less sure of that last part than I was a second ago
 
Jim
yes, but my main concern is can we transform $A$ like $B$ as it is given that
$A$ is another permutation of $B$. i am trying to do that using $H$. it is somewhat 'isomorphism testing problem'.
 
@Jim The matrices are not actually relevant. This is purely a question inside $S_n$
 
Jim
@TobiasKildetoft I encountered the problem there, as I am going to use it there, I included matrix in my post.
 
@Jim Actually, it seems like we can never be sure of more than one fixed point (take the permutation sending $i$ to $n-i+1$).
Anyway, I need to go teach now.
 
Jim
11:58 AM
@TobiasKildetoft please, thanks for your time.
@ThomasKlimpel , Hi.
3
Q: Is there any efficient algorithm for finding subgroups of a given finite group?

AnalysisI am implementing an algorithm which finds every subgroup of given group. Here's my algorithm. Let $G$ be a group of order $n$ with elements $g_1,\cdots,g_n$. Then I consider each $\langle g_i\rangle $ which are all subgroups of $G$. Of course, if $\langle g_i\rangle =\langle g_j\rangle $, we...

 
12:30 PM
@Jim Hi, I will look at those questions later...
 
Jim
@ThomasKlimpel take your time.
1st question is more important(Sylow Subgroup of a Symmetric Group acting on Non-Symmetric Matrix) to me than second ,
 

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