Basic properties of blocks of a group action

Let G \leq S_A act transitively on the set A. A block is a nonempty subset B \subseteq A such that for all \sigma \in G either \sigma[B] = B or \sigma[B] \cap B = \emptyset.

  1. Prove that if B is a block containing a \in A and we define \mathsf{stab}(B) = \{ \sigma \in G \ |\ \sigma[B] = B \} then \mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G.
  2. Prove that if B is a block then \mathcal{B} = \{ \sigma[B] \ |\ \sigma \in G \} is a partition of A.
  3. A (transitive) group G \leq S_A is called primitive if the only blocks in A are the trivial ones- the singletons and A itself. Prove that S_4 is primitive on A = \{1,2,3,4\}. Prove that D_8 is not primitive as a permutation group on the four vertices of a square.
  4. Let G \leq S_A act transitively on A. Prove that G is primitive on A if and only if for all a \in A, \mathsf{stab}(a) is maximal in G.

  1. Note that \mathsf{stab}(B) is not empty since 1[B] = B. Now let \sigma, \tau \in \mathsf{stab}(B). Note that \tau^{-1}[B] = \tau^{-1}\tau [B] = B, so that \sigma\tau^{-1}[B] = B. Thus \sigma\tau^{-1} \in \mathsf{stab}(B). By the subgroup criterion, \mathsf{stab}(B) \leq G.

    Now suppose \sigma \in \mathsf{stab}(a). We have \sigma(a) = a, so that \{a\} \subseteq \sigma[B] \cap B. Hence \sigma[B] \cap B is nonempty, and we have \sigma[B] = B since B is a block. Thus \sigma \in \mathsf{stab}(B). By a previous exercise, \mathsf{stab}(a) \leq \mathsf{stab}(B).

  2. First we show that \bigcup_{\sigma \in G} \sigma[B] = A. The (\subseteq) direction is clear. (\supseteq): Let a \in A and b \in B. Since the action of G is transitive, there exists \sigma \in G such that a = \sigma(b). Then a \in \sigma[B], so that aA \subseteq \bigcup_{\sigma \in G} \sigma[B].

    Now suppose \sigma[B] \cap \tau[B] \neq \emptyset. Then there exist a,b \in B such that \sigma(a) = \tau(b). Now a = \sigma^{-1}\tau \cdot b, so that \sigma^{-1}\tau[B] \cap B is not empty. Thus \sigma^{-1}\tau[B] = B, and we have \sigma[B] = \tau[B]. So elements of \mathcal{B} are pairwise disjoint; hence \mathcal{B} is a partition of A.

    1. Let B \subseteq A = \{1,2,3,4\} be a proper nonempty subset. Then there exist elements x \in B and y \in A \setminus B. Suppose B is a block. Consider \sigma = (x\ y) \in S_4; since \sigma(x) \notin B, we have \sigma[B] \cap B = \emptyset. Let w and z be the remaining elements of A. If w \in B, then w \in \sigma[B] \cap B, a contradiction; similarly for z. Thus B = \{x\}. Now clearly A itself is a block, and any proper block must be a singleton. Thus this action of S_4 is primitive.
    2. We saw previously that D_8 \cong \langle (1\ 3), (1\ 2\ 3\ 4) \rangle, where A = \{1,2,3,4\} are labels on the vertices of a square (written clockwise). Consider the set \{1,3\}. We see that
      1. 1 \cdot \{1,3\} = \{1,3\}
      2. (1\ 2\ 3\ 4) \cdot \{1,3\} = \{2,4\}
      3. (1\ 3)(2\ 4) \cdot \{1,3\} = \{1,3\}
      4. (1\ 4\ 3\ 2) \cdot \{1,3\} = \{2,4\}
      5. (1\ 3) \cdot \{1,3\} = \{1,3\}
      6. (1\ 2)(3\ 4) \cdot \{1,3\} = \{2,4\}
      7. (2\ 4) \cdot \{1,3\} = \{1,3\}
      8. (1\ 4)(2\ 3) \cdot \{1,3\} = \{2,4\}

      So that \{1,3\} is a nontrivial block; hence this action is not primitive.

  3. We begin with some lemmas.

    Lemma 1: Let G \leq S_A act transitively on A and let B \subseteq A be a block under the action. Then \mathsf{stab}(B) = G if and only if B = A. Proof: The (\Leftarrow) direction is clear. (\Rightarrow) Suppose B is a proper subset and let x \in B, y \in A \setminus B. Now y \in (x\ y)[B] and y \notin B, so that (x\ y)[B] \neq B. Since B is a block, we have (x\ y)[B] \cap B = \emptyset. But x \in (x\ y)[B] \cap B since x \in B and y \notin B, a contradiction. So B is not proper, and we have A = B. \square

    Now we move to the main result.

    (\Rightarrow) Suppose G is primitive on A; then the only blocks of A are A and the singletons. Now let a \in A and let H \leq G be a subgroup with \mathsf{stab}(a) \leq H \leq G. We consider the set H \cdot a = \{ h \cdot a \ |\ h \in H \}.

    We claim that H \cdot a is a block. Proof of claim: H \cdot a is not empty since a = 1 \cdot a \in H \cdot a. Now let \sigma \in G. If \sigma \in H, then \sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a. If \sigma \notin H, then \sigma \cdot (H \cdot a) = \sigma H \cdot a. Suppose that \sigma H \cdot a \cap H \cdot a \neq \emptyset; say that \sigma\tau_1 \cdot a = \tau_2 \cdot a for some \tau_1, \tau_2 \in H. Then \tau_2^{-1} \sigma \tau_1 \cdot a = a, so that \tau_2^{-1} \sigma \tau_1 \in \mathsf{stab}(a) \leq H. But then \sigma \in \tau_2H\tau_1^{-1} = H, a contradiction. Thus if \sigma \notin H, then \sigma[H \cdot a] \cap H \cdot a = \emptyset. Hence H \cdot a is a block.

    Next we claim that \mathsf{stab}(H \cdot a) = H. Proof of claim: (\supseteq) If \tau \in H, then \tau \cdot (H \cdot a) = \tau H \cdot a = H \cdot a. (\subseteq) Suppose \sigma \cdot (H \cdot a) = \sigma H \cdot a = H \cdot a. Then \sigma \cdot a = \tau \cdot a for some \tau \in H, hence \tau^{-1} \sigma \cdot a = a. Then \tau^{-1} \sigma \in \mathsf{stab}(a) \leq H, so that \sigma \in H.

    Since G is primitive on a and a \in H \cdot a, we have H \cdot a = \{a\} or H \cdot a = A. If H \cdot a = \{a\}, we have H \leq \mathsf{stab}(a), so that H = \mathsf{stab}(a). If H \cdot a = A, we have H = \mathsf{stab}(H \cdot a) = \mathsf{stab}(A) = G. Thus \mathsf{stab}(a) is a maximal subgroup of G.

    (\Leftarrow) Suppose that for all a \in A, \mathsf{stab}(a) is a maximal subgroup in G. Let B \subseteq A be a block with a \in B. Now \mathsf{stab}(a) \leq \mathsf{stab}(B) \leq G by part 1 above. Since \mathsf{stab}(a) is maximal, there are two cases.

    If \mathsf{stab}(B) = \mathsf{stab}(a) and b \in B such that b \neq a, then since G acts transitively there exists \sigma \in G such that \sigma \cdot a = b. Now \sigma \in \mathsf{stab}(B) = \mathsf{stab}(a), a contradiction. Thus B = \{a\}.

    If \mathsf{stab}(B) = G, then by Lemma 1 we have B = A.

    Thus G is primitive on A.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: