Why is D4 not primitive on the vertices of a square?

[fishturtle1]

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).

 

[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 $\lbrace a, b, c, d \rbrace$. []