I was thinking about the problem from Dummit-Foote (Art. 4.1 Question 9)enter image description here

N.B. The definition of block as given in Ex. 7: Let $G$ be a transitive permutation group on the finite set $A$. A block is a nonempty subset $B$ of $A$ such that for all $\sigma\in G$ either $\sigma(B) = B$ or $\sigma(B)\cap B = \emptyset$.

I'm not looking for a solution. I did it for the first part. What I want to know is that why $O_i$'s are blocks? $G$ is not necessarily a permutation group. Do I need to consider the effect of the group $\{\sigma_g:A\to A:a\mapsto ga\}$ instead of $G?$

Please clarify and help me. By the way I would like to have an elaborate answer as I'm self-studying Group action for the first time.


Given an action of a group $G$ on a set $A$, we can define the permutation representation of $G$. More specifically, define $\rho:G\to S_A$ by $\rho(g)(a)=g\cdot a$. This map might not be injective, but if the action is transitive, then $\operatorname{im}G$ is now a transitive permutation group on the finite set $A$.

Now, define the blocks of $G$ acting on $A$ to be those of the transitive permutation group $\operatorname{im}G$ acting on $A$ (using the definition you've listed). With this definition, it follows that the $\mathcal{O}_i$ are blocks under the action of $G$, because they are blocks under the action of $\operatorname{im}G$ with your original definition.

  • $\begingroup$ Ahh...!! I made a correct interpretation. $\endgroup$ – Sriti Mallick Sep 6 '13 at 16:51
  • $\begingroup$ @SritiMallick: You did. I'm sorry I didn't notice that before. $\endgroup$ – Jared Sep 6 '13 at 16:55

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