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I have a basic question in algebra.

Question:

Let $n$ be a non-negative integer. For any family $ (i_1, \ldots, i_r) $ of non-negative integers such that $ i_1 + \ldots + i_r = n $, we denote by $ D(n, i_1, \ldots, i_r) $ the set of families $ (I_1, \ldots, I_r) $ of pairwise disjoint subsets of $ \left \{1;2;...;n \right \} $ such that the cardinality of each $ I_\ell $ is $ i_\ell $ for all $ \ell $.
For any family $ (i_1, \ldots, i_r) $ of integers $ \geq 0 $ such that $ i_1 + \ldots + i_r = n $, briefly describe a natural operation of $ S_n $ on $ D(n, i_1, \ldots, i_r) $, and show that it is transitive.

My answer

1-
First of all let try to undersand how does look like $D(n, i_1, \ldots, i_r)$.
In order to do that we take $ (1,2) \Rightarrow n=3, r=2 $. So a collection of $ (I_1, I_2) $ of pairwise disjoint subsets of $ {1, 2,3} $ can be $(I_1=\left \{ 2 \right \}, I_2= \left \{ 1;3 \right \})$. In fact there are other ones collection of $(I_1, I_2)$ that satisfies all the criterium. $D(n, i_1, \ldots, i_r) = \left \{ (I_1=\left \{ 1 \right \}, I_2= \left \{ 2;3 \right \}) ; (I_1=\left \{ 2 \right \}, I_2= \left \{ 1;3 \right \}) ; (I_1=\left \{ 3 \right \}, I_2= \left \{ 1;2 \right \}) \right \} $

2-
$\forall \sigma \in S_n$ is a bijective application from $\left \{1;2;...;n \right \}$ to itself. and because $\forall (I_1, \ldots, I_r) \in D(n, i_1, \ldots, i_r)$ the "$I_l$" form a partition of $\left \{1;2;...;n \right \}$ we get that $\sigma (I_1, \ldots, I_r) = (\sigma(I_1), \ldots, \sigma(I_r))$. In words $\sigma(.)$ acts on this family by permuting the elements of each subset $I_l$ according to $\sigma(.)$.

After I have difficulties to answer and to show rigoursly the last part (conscerning the transitivity).
Moreover if you have any remarks on what I ve written in "1-" and "2-" I will be happy to read them.

Thank for your help.

citadel
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OffHakhol
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    Proving transitivity of the action is not difficult. The main problem is awkward notation, which is difficult to avoid. I will demonstrate by an example, and leave you to try and write it out formally. Let's take $n=7$, $i_1,i_2,i_3=2,1,4$. Now choose two "random" elements of $D(7,2,1,4)$, say $A= ({3,6},{4},{1,2,5,7})$ and $B = ({5,7},{1}, {2,3,4,6})$. Then $\sigma \in S_7$ mapping $1,2,3,4,5,6,7$ to $2,3,5,1,4,7,6$ maps $A$ to $B$. Note that $\sigma$ in not unique - in fact there are $2! \times 1! \times 4!$ possible $\sigma$. – Derek Holt Dec 29 '23 at 19:11
  • @DerekHolt I think that what is complicate for me is to understand what it means here to be "transitive" I know the definition of transitivity but what is the relation her on which we can apply it? – OffHakhol Dec 31 '23 at 15:27
  • @DerekHolt I ve written an answer I will be happy to have your feedback on it. – OffHakhol Dec 31 '23 at 16:50

1 Answers1

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The main problem in my understanding/ solving of this question is that I didn't know yet that we were talking about transitivity in a group. Indeed I thought that we were talking on the transitive relation. That's why now my question seems to me very basic. :'-) .

Based on the comment of https://math.stackexchange.com/users/2820/derek-holt

1-
Let first take an exemple in order to understand.
We take $n=7; i_1 , i_2 , i_3 = 2,1,4$. Now choose two "random" elements of $D(7; 2,1,4)$. For exemple $A=( \left \{ 3;6 \right \}, \left \{ 4 \right \}, \left \{1;2;5;7 \right \} )$ and $B=(\left \{ 5;7 \right \}, \left \{ 1 \right \}, \left \{2;3;4;6 \right \})$. Then $\sigma \in S_7$ can for exemple maps $\left \{1,2,3,4,5,6,7 \right \}$ to $ \left \{ 2,3,5,1,4,7,6 \right \} $ that means that $\sigma '$ maps $A$ to $B$.
More precisally $I_{1;A}=\left \{3; 6 \right \}$ and $ I_{1;B}=\left \{5; 7 \right \}$. So we define $ 3 \rightarrow 5$ , $6 \rightarrow 7$...etc...
Those here $S_7$ is our group and $\sigma$ is our "group action" on the set $D(7; 2,1,4)$. In order to prove the transitivity we then need to show that $\forall I_x, I_y \in D(7; 2,1,4) \Rightarrow \sigma \in S_7 $ s.t. $I_y = \sigma(I_x)$ and this is precisally waht we ve found here, a $\sigma \in S_7 $ that links $A$ and $B$.

2-
Now more generally let's take two specific $I_x, I_y \in D(n; i_1,i_2,...,i_r) \Rightarrow I'$ is of the form $I_x=(I_{1;x};...;I_{r;x}), I_y=(I_{1;y};...;I_{r;y})$. We need to prove that $\exists \sigma \in S_n$ s.t. $\sigma(I_x)= I_y$.
It is very easy to do that what we need to do is for exemple to index the elements of $I_{1;x}$ from $1$ to $i_1$, those of $I_2$ from $i_1+1$ to $i_1+i_2$, etc...
So $I_x$ looks like this $ I_x = \left\{1;2;...;i_1;i_1+1;...;i_1+i_2;i_1+i_2+1;...;i_1+...+i_r \right \}=\left\{1;2;...;i_1;i_1+1;...;n \right \}$ .
Proceed the same way for $I_y$ but instead we use the index $j$, hence $I_y$ looks like this $ I_y = \left\{1;2;...;j_1;j_1+1;...;j_1+j_2;j_1+j_2+1;...;j_1+...+j_r \right \}=\left\{1;2;...;j_1;j_1+1;...;n \right \}$
Now it is easy to construct a bijection from $I_x$ to $I_y$ as follow $\forall 1 \leq i \leq n \Rightarrow \sigma(i)=j=i$ that means that for exemple $\sigma$ maps the element indexed by $i_1$ to the element indexed by $j_1$.
And of course as all the elements of different indexes correspond to different numbers so $\sigma$ is a bijection in $S_n$.

OffHakhol
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    This is not completely formal, but it is essentially correct. As I said, the main problem is battling with the notation. In general the notion of transitivity applies to an action of a group $G$ on a set $X$. The action is transitive if for all $x,y \in X$ there exists $g \in G$ with $g \cdot x = y$. – Derek Holt Dec 31 '23 at 17:28
  • @DerekHolt thank for your feedback. The definition you ve writtent this exactly is what I ve tried to prove. – OffHakhol Dec 31 '23 at 17:30