2
$\begingroup$

I want to solve the following exercise from Dummit & Foote's Abstract Algebra text:

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\} $).

(a) Prove that if $B$ is a block containing the element $a$ of $A$, then the set $G_B$ defined by $G_B=\{\sigma \in G\, | \, \sigma(B)=B\}$ is a subgroup of $G$ containing $G_a$.

(b) Show that if $B$ is a block and $\sigma_1(B),\sigma_2(B),\dots,\sigma_n(B)$ are all the distinct images of $B$ under the elements of $G$, then these form a partition of $A$.

(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 $S_4$ is primitive on $A=\{1,2,3,4\}$. Show that $D_8$ is not primitive as a permutation group on the four vertices of a square

(d) Prove that the transitive group $G$ is primitive on $A$ if and only if for each $a \in A$, the only subgroups of $G$ containing $G_a$ are $G_a$ and $G$ (i.e., $G_a$ is a maximal subgroup of $G$, cf. Exercise 16, Section 2.4). [Use part (a).]


This is my attempt:

(a) Suppose $\sigma,\tau \in G_B$, the equality $\sigma(B)=B$ gives $B=\sigma^{-1}(B)$, and acting on this by $\tau$ gives \begin{equation} B=\tau(B)=\tau(\sigma^{-1}(B))=(\tau \circ \sigma^{-1})(B), \end{equation} so $G_B \leq G$. Moreover, if $\sigma \in G_a$ fixes $a \in B$ then $\sigma(B) \cap B \supseteq \{a\}$. Since $B$ is a block the only option is $\sigma(B)=B$. This gives $G_a \subseteq G_B$.

(b) We first prove that the images $\sigma_i(B)$ are disjoint. For suppose $\sigma_i(B) \cap \sigma_j(B)$ is nonempty for $i \neq j$. Acting on this with $\sigma_i^{-1}$ gives that the intersection $B \cap (\sigma_i^{-1} \circ \sigma_j)(B)$ is nonempty. Since $B$ is a block we must have that $(\sigma_i^{-1} \circ \sigma_j)(B)=B$. Acting on this with $\sigma_i$ gives $\sigma_i(B)=\sigma_j(B)$ which is a contradiction, since the images are known to be distinct.

We also need to show that the union of these images is $A$. Suppose by way of contradiction that there is some $a \in A$ with $a \notin \bigcup_{i=1}^n \sigma_i(B)=\bigcup_{\sigma \in G} \sigma(B)$. Since $G$ is transitive, we can find a permutation sending some $b \in B$ to $a$. It follows that $\sigma_0(B) \supseteq \sigma_0(\{b\})=\{a\}$, which is a contradiction.

(c) We prove that there are no blocks $B$ of order 2. Suppose $B=\{i,j\}$ is such a block (and the remaining elements are $k,l$). The permutation $(i \; k)$ shows $B$ is not a block. We also prove that there are no blocks of order 3. Suppose $B=\{i,j,k\}$ is such a block. The permutation $(i \; l)$ shows that $B$ is not a block. Thus the only possible orders for a block $B \subseteq A=\{1,2,3,4\}$ are 1 or 4, and $S_4$ is primitive on $A$.

Since $D_8$ acts on the set of unordered pairs of opposite vertices, we have that such a pair (for example $B=\{1,3\}$) is a block. This is because for any $g \in D_8$ we either have $g \cdot \{1,3\}=\{1,3\}=B$ or $g \cdot \{1,3\}=\{2,4\}$ is disjoint with $B$.

(d) We prove both implications:

  • Suppose for each $a \in A$, the only subgroups of $G$ containing $G_a$ are $G_a$ and $G$, and let $B \subseteq A$ be a block, and let $b$ be a point in that block. Part (a) establishes $G_B$ as a supergroup of $G_b$, and hence gives two possible cases:

    1. $G_B=G$. In that case, we claim that $B=A$, for if $B \subsetneq A$ there is some $b' \in B \setminus A$. Since $G$ is transitive, there is some permutation $\sigma \in G$ sending $b$ to $b'$. But that permutation doesn't fix $B$, which is a contradiction.

    2. $G_B=G_b$. In that case I don't know what to do...

  • The reverse implication should be here someday...


I now have several questions:

  1. Is my proof of parts (a),(b),(c) correct?

  2. Is it necessary that the set $A$ is finite? So far I didn't have to use that even once.

  3. Could anyone please help me fill in the rest? I found a solution here, but the third section from the end seems flawed to me...

Thank you!

$\endgroup$
1
$\begingroup$

I think the link you provided proves the last part correctly. The key is recognizing the correspondence between block and subgroups. Fixing $a \in A$, then there is a correspondence between \begin{align*} \{\text{subgroups } H \text{ with } G_a \subseteq H \subseteq G \} & \longleftrightarrow \{\text{subsets } B \text{ with } \{a\} \subseteq B \subseteq A\}\\ H &\longmapsto \{\sigma(a) : \sigma \in H\}\\ G_B = \{\sigma \in G : \sigma(B) = B\} \ &{\longleftarrow\!\shortmid} \ B \end{align*} which is proven in your link. We can prove part (d) using this fact.

($\Leftarrow$): To complete your argument, note that since the blocks partition $A$, we can choose a block $B$ that contains $a$. You already covered the case $G_B = G$, so suppose $G_B = G_a$. For contradiction, assume $B \neq \{a\}$ so that there is some $b \neq a$ with $b \in B$. Since $G$ is transitive, then there exists $\sigma \in G$ such that $\sigma(a) = b$. Then $b \in \sigma(B) \cap B$, so $\sigma(B) = B$. Then $\sigma \in G_B$ but $\sigma \notin G_a$, contradiction.

($\Rightarrow$): Suppose that $G$ is primitive and there is a subgroup $H$ with $G_a \subseteq H \subseteq G$. Let $B = \{\sigma(a) : \sigma \in H\}$. Since $G$ is primitive, then either $B = \{a\}$ or $B = G$. In the first case ($B = \{a\}$), then every element of $H$ stabilizes $a$, so $H = G_a$. In the latter case ($B = G$), then $H = G_B = G_A = G$ since the link proved $H = G_B$.

$\endgroup$
  • $\begingroup$ I think you should add to the $(\Leftarrow)$ part that $\sigma(B)=B$ because the intersection is nonempty. Also, it seems that the condition that $A$ is finite is unnecessary after all... $\endgroup$ – user1337 Jun 24 '14 at 12:45
  • $\begingroup$ @user1337 Okay, I'll add that. I think the finiteness is used in the assumption in part (b) that there are only finitely many distinct images of $B$ under $G$. $\endgroup$ – André 3000 Jun 24 '14 at 13:39
  • $\begingroup$ Thank you. Would you agree with me that having infinitely many images is not an essential issue? $\endgroup$ – user1337 Jun 24 '14 at 14:03
  • $\begingroup$ I'll have to read through the problem carefully to be sure that $A$ being finite isn't used anywhere else, but from a first look, it doesn't seem to be. $\endgroup$ – André 3000 Jun 24 '14 at 20:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.