Transitive permutation groups and blocks (D&F 4.1.7)

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 \(\sigma(B) \cap B = \varnothing\).

(a) Prove that if \(B\) is a block containing \(a \in A\), then \(G_B\) is a subgroup of \(G\) containing \(G_a\).

Trivial.

(b) Show that if \(B\) is a block, then the set of distinct images of \(B\) under the elements of \(G\) \(\{\sigma_1(B), \sigma_2(B), \ldots, \sigma_n(B)\}\) is a partition of \(A\).

First we show that \(A = \bigcup_{\sigma \in G} \sigma(B)\). Fix a point \(b \in B\) and let \(a \in A\) be arbitrary. Since \(G\) is transitive, there is a permutation \(\sigma\) such that \(a = \sigma(b) \in \sigma(B)\). Thus \(A \subseteq\bigcup_{\sigma \in G} \sigma(B)\); the other direction is obvious.

Next we show that for distinct permutations \(\sigma\) and \(\tau\), either \(\sigma(B) = \tau(B)\) or \(\sigma(B) \cap \tau(B) = \varnothing\). Suppose that neither is true. Then there is some \(a \in A\) which is in both \(\sigma(B)\) and \(\tau(B)\), and also some \(b \in A\) which is in \(\sigma(B)\) but not \(\tau(B)\). So \(\sigma^{-1}(a), \sigma^{-1}(b), \tau^{-1}(a) \in B\) but \(\tau^{-1}(b) \not\in B\).

Now consider the permutation \(\tau^{-1}\sigma \in G\), which maps \(\sigma^{-1}(a) \in B\) to \(\tau^{-1}(a) \in B\) and \(\sigma^{-1}(b) \in B\) to \(\tau^{-1}(b) \not\in B\). This means that \(B\) cannot be a block, because \(\tau^{-1}\sigma(B)\) is neither equal to \(B\) nor disjoint from \(B\). Hence \(\sigma(B)\) and \(\tau(B)\) are either equal or disjoint, and the conclusion follows.

(c) A transitive group \(G\) on a set \(A\) is primitive if the only blocks in \(A\) are the singletons and \(A\) itself. Show that \(S_4\) is primitive on \(A = \{1,2,3,4\}\) and that \(D_8\) is not primitive as a permutation group on the four vertices of a square.

\(S_4\): Let \(B\) be a block and \(n = \left|B\right|\). Clearly the distinct images of \(B\) under the elements of \(G\) are precisely the \(n\)-element subsets of \(A\). These only form a partition when \(n = 1\) or \(\left|A\right|\).

\(D_8\): Consider the subset \(B = \{1,3\}\), where the standard labelling of vertices applies. It can be verified that all permutations in \(D_8\) send \(B\) to either itself or \(\{2,4\}\), a disjoint set. Thus \(B\) is a block and so \(D_8\) is not primitive.

(d) Prove that the transitive group \(G\) is primitive on \(A\) if and only if for each \(a \in A\), \(G_a\) is a maximal subgroup of \(G\).

\(\Leftarrow\): Let \(B\) be a block containing \(a\). Then either \(G_B = G_a\) or \(G_B = G\). If the former case is true, then for all \(\sigma \in G\), \(\sigma(a) \in B\) if and only if \(\sigma(a) \in \{a\}\). Since \(G\) is transitive, for all \(b \in B\) there is some \(\sigma_b\) mapping \(a\) to \(b\), so \(B = \{a\}\). If the latter case is true, then for all \(\sigma \in G\) and \(b \in B\), \(\sigma(b) \in B\). By transitivity of \(G\), we also have that for all \(a \in A\), there is some \(\sigma_a\) such that \(a = \sigma_a(b) \in B\). So \(B = A\).

\(\Rightarrow\): We show that for any subgroup \(H\) of \(G\) containing \(G_a\), the set \(Ha = \{h \cdot a \mid h \in H\}\) is a block containing \(a\), and \(G_{Ha} = H\). For all \(h' \in H\) and \(h \cdot a \in Ha\), \(h' \cdot (h \cdot a) = (h'h) \cdot a \in Ha\). Furthermore, \(h'\) maps injectively (and hence surjectively) from \(Ha\) to itself, i.e. \(\sigma_{h'}(Ha) = Ha\).  On the other hand, for all \(g \in G-H\), \(g \cdot (h \cdot a) = (gh) \cdot a \not\in Ha\), otherwise \(g\) would be in \(H\). In other words \(\sigma_{g}(Ha) \cap Ha = \varnothing\). Furthermore, \(x \in G_{Ha}\) iff \(x \cdot (h \cdot a) = (xh) \cdot a \in Ha\) for all \(h \cdot a\in Ha\), iff \(x \in H\).

Based on the above result and the fact that \(G\) is primitive, it is evident that each subgroup \(H\) containing \(G_a\) is either \(G_a\) or \(G_A = G\).