I 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 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 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
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?
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
@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
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
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'.
I 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...