# Group action and blocks

Homework Statement:
Let ##G## be a transitive permutation group on the finite set ##A##. A block is a nonempty subset of ##B## of ##A## such that for all ##\rho \in G## either ##\rho(B) = B## or ##\rho(B) \cap B = \emptyset##. (here ##\rho(B) = \lbrace \rho(b) : b \in B \rbrace##).

c) A (transitive) group ##G## acts on a set ##A## is set to be primitive if the only blocks in ##A## are the trivial ones: the sets of size ##1## and ##A## itself. Show that ##S_4## is primitive on ##A = \lbrace 1, 2, 3, 4, \rbrace##. Show that ##D_8## is not primitive as a permutation group of on the four vertices of a square.
Relevant Equations:
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Proof: Let ##B = \lbrace a \rbrace \subseteq A## and ##\rho \in S_4##. We have two cases, ##\rho(a) = a## in which case ##\rho(B) = B##, or ##\rho(a) \neq a## in which case ##\rho(B) \cap B = \emptyset##. Its clear that ##\rho(A) = A##. So these sets are indeed blocks. Now let ##C## be any subset of ##A## such that ##2 \le \vert C \vert \le 3##. Then there is ##x, y \in C, z \not\in C## and ##\gamma \in S_4## such that ##\gamma(x) = z## and ##\gamma(y) = y##. This implies ##\gamma(C) \neq C## and ##\gamma(C) \cap C \neq \emptyset##. So ##C## is not a block. We can conclude ##S_4## is primitive on ##A##. []

For the second part we need to show ##D_8## is not primitive on the set of vertices of a square. We label the vertices ##1, 2, 3, 4##. Let ##S \subseteq A## such that ##\vert S \vert = 2## or ##3##. If ##\vert S \vert = 3## and ##\sigma_1 = (1234)##, then ##\sigma_1(S) \cap S \neq \emptyset## and ##\sigma_1(S) \neq S##. So, suppose ##\vert S \vert = 2##. Then ##S = \lbrace a, b \rbrace## and there is ##c \not\in S##. Let ##\sigma_2 = (ac)##. Then ##\sigma_2(S) = \lbrace c, b \rbrace##. It follows ##S## is not a block. Since ##D_8 \subseteq S_4##, it follows the only blocks of ##D_8## acting on ##A## are the trivial ones, and so ##D_8## is primitive on ##A##.

Where did I go wrong in the second part?

member 587159
Homework Statement:: Let ##G## be a transitive permutation group on the finite set ##A##. A block is a nonempty subset of ##B## of ##A## such that for all ##\rho \in G## either ##\rho(B) = B## or ##\rho(B) \cap B = \emptyset##. (here ##\rho(B) = \lbrace \rho(b) : b \in B \rbrace##).

c) A (transitive) group ##G## acts on a set ##A## is set to be primitive if the only blocks in ##A## are the trivial ones: the sets of size ##1## and ##A## itself. Show that ##S_4## is primitive on ##A = \lbrace 1, 2, 3, 4, \rbrace##. Show that ##D_8## is not primitive as a permutation group of on the four vertices of a square.
Homework Equations:: .

Proof: Let ##B = \lbrace a \rbrace \subseteq A## and ##\rho \in S_4##. We have two cases, ##\rho(a) = a## in which case ##\rho(B) = B##, or ##\rho(a) \neq a## in which case ##\rho(B) \cap B = \emptyset##. Its clear that ##\rho(A) = A##. So these sets are indeed blocks. Now let ##C## be any subset of ##A## such that ##2 \le \vert C \vert \le 3##. Then there is ##x, y \in C, z \not\in C## and ##\gamma \in S_4## such that ##\gamma(x) = z## and ##\gamma(y) = y##. This implies ##\gamma(C) \neq C## and ##\gamma(C) \cap C \neq \emptyset##. So ##C## is not a block. We can conclude ##S_4## is primitive on ##A##. []

For the second part we need to show ##D_8## is not primitive on the set of vertices of a square. We label the vertices ##1, 2, 3, 4##. Let ##S \subseteq A## such that ##\vert S \vert = 2## or ##3##. If ##\vert S \vert = 3## and ##\sigma_1 = (1234)##, then ##\sigma_1(S) \cap S \neq \emptyset## and ##\sigma_1(S) \neq S##. So, suppose ##\vert S \vert = 2##. Then ##S = \lbrace a, b \rbrace## and there is ##c \not\in S##. Let ##\sigma_2 = (ac)##. Then ##\sigma_2(S) = \lbrace c, b \rbrace##. It follows ##S## is not a block. Since ##D_8 \subseteq S_4##, it follows the only blocks of ##D_8## acting on ##A## are the trivial ones, and so ##D_8## is primitive on ##A##.

Where did I go wrong in the second part?

A possible problem I see is the following:

How are you sure that ##\sigma_2\in D_8##? Not every transposition of ##S_4## is in ##D_8## (because ##S_4## is generated by all its transpositions).

Last edited by a moderator:
fishturtle1
A possible problem I see is the following:

How are you sure that ##\sigma_2\in D_8##? Not every transposition of ##S_4## is in ##D_8## (because ##S_4## is generated by all its transpositions).
Thanks!

Consider the square with vertices ##a = (0,1), b = (1, 1), c = (1,0), d = (0,0)##. Let ##S = \lbrace a, c \rbrace##. Observe,
##(abcd)\cdot S = (abcd)^3\cdot S = (ab)(dc)\cdot S = (ad)(bc)\cdot S = \lbrace b, d \rbrace##
and
##1\cdot S = (abcd)^2 \cdot S = (ac) \cdot S = bd\cdot S = S##.
This shows that for all ##\sigma \in D_4##, we have ##\sigma(S) = S## or ##\sigma(S) \cap S = \emptyset##. This means that ##S## is a nontrivial block. So ##D_4## is not primitive on ##\lbrace a, b, c, d \rbrace##. []