Im trying to solve this problem from Dummit & Foote:

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 $\sigma(B)\cap B=\emptyset$ (here $\sigma(B)$ is the set $\{\sigma(b)|b\in B\}$).

(c) A (transitive) group G on a set A is said to be primitive if the only blocks in A are the trivial ones: the sets of size 1 and A itself. Show that S4 is primitive on A={1,2,3,4}. Show that D8 is not primitive as a permutation group on the four vertices of a square

In part (c) i have to show that the subsets are {1,2} and {1,2,3} are not blocks, but im confused with the operation that i need to perform, namely:



I know that the answers are: (13){1,2} = {1,3} and (34){1,2,3} = {1,2,4}, but i don't understand the procedure.

  • $\begingroup$ I think that the answer should be $(13)\{1,2\}=\{2,3\}$. $\endgroup$ – Mike Earnest Sep 27 '14 at 0:59

Let's say you have a cycle $(a_1 a_2 a_3 ... a_n)$. The action of this cycle on a set of numbers is, by definition, the one that sends $a_1$ to $a_2$, $a_2$ to $a_3$, $a_3$ to $a_4$ and so on, until finally $a_n$ to $a_1$. So that means for example if you have $(34)$ acting on the set $\{1,2,3\}$, it just sends $3$ to $4$ (because there's no $4$ to send to $3$,) so

$$ (34)\{1,2,3\}=\{1,2,4\}.$$

For another example,


because $2$ goes to $3$ and $5$ goes to $2$. The rest aren't changed.


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