Plinth theorem

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Statement

Suppose G is a finite group with a faithful group action on a set S. Suppose N_1 and N_2 are two distinct minimal normal subgroups of G that act transitively on S under the induced action. Then, the following are true:

  1. C_G(N_1) = N_2 and C_G(N_2) = N_1.
  2. N_1 is isomorphic to N_2.
  3. There exists an element \sigma \in \operatorname{Sym}(S) such that \sigma is involutive, \sigma normalizes G, and \sigma conjugates N_1 to N_2.
  4. The action of G on S is a primitive group action.

Related facts

Proof

Given: A finite group G with a faithful group action on a set S. N_1 and N_2 are two distinct minimal normal subgroups of G that act transitively on S under the induced action.

To prove:

  1. C_G(N_1) = N_2 and C_G(N_2) = N_1.
  2. N_1 is isomorphic to N_2.
  3. There exists an element \sigma \in \operatorname{Sym}(S) such that \sigma is involutive, \sigma normalizes G, and \sigma conjugates N_1 to N_2.
  4. The action of G on S is a primitive group action.

Proof:

  1. N_1 and N_2 intersect trivially: Since an intersection of normal subgroups is normal (Fact (1)), and N_1 and N_2 are distinct minimal normal subgroups, so their intersection must be trivial.
  2. Every element of N_1 commutes with every element of N_2: Since N_1 and N_2 are normal subgroups, the commutator [N_1,N_2] is contained in the intersection N_1 \cap N_2. Thus, [N_1,N_2] is trivial, and thus, every element of N_1 commutes with every element of N_2.
  3. The induced action of N_1 on S is regular: Suppose there exists a non-identity element g \in N_1 and s \in S such that g \cdot s = s. Now, for every h \in N_2, we have gh = hg, so (gh) \cdot s = (hg) \cdot s, so g fixes h \cdot s. Since N_2 is transitive on S, we obtain that g acts as the identity on S, a contradiction to the assumption that the action was faithful. Thus, the action of N_1 on S is regular.
  4. The induced action of N_2 on S is regular: This follows from the same reasoning as the previous step, interchanging the roles of N_1 and N_2.
  5. The centralizer of N_1 in \operatorname{Sym}(S) is isomorphic to N_1: Let s_0 \in S be any chosen element. This can be seen in two ways:
    • Suppose f \in \operatorname{Sym}(S) commutes with the action of N_1. Then, we must have f(g \cdot s_0) = g \cdot (f(s_0)) for all g \in N_1. Thus, f is completely determined by the value of f(s_0). Further, if f_1(s_0) = h_1 \cdot s_0 and f_2(s_0) = h_2 \cdot s_0, then (f_2 \circ f_1) (s_0) = f_2(h_1 \cdot s_0) = h_1 \cdot h_2 \cdot s_0 = (h_1h_2) \cdot s_0. Thus, there is a bijection between the centralizer of N_1 in \operatorname{Sym}(S) and the elements of N_1, that reverses the order of multiplication. By fact (2) (every group is isomorphic to its opposite group), we obtain that the centralizer of N_1 in \operatorname{Sym}(S) is isomorphic to N_1.
    • A shorter way is to observe that the action of N_1 on S is equivalent to the left-regular group action on itself. Now, we use the fact that the only permutations that commute with all left multiplications are the right multiplications, which form the right-regular group action. Thus, the centralizer of N_1 in \operatorname{Sym}(S) is an isomorphic group acting via right multiplication.
  6. N_2 is a subgroup of the centralizer of N_1 in \operatorname{Sym}(S), hence isomorphic to a subgroup of N_1: This combines step (2) with the previous step.
  7. N_1 is a subgroup of the centralizer of N_2 in \operatorname{Sym}(S), hence isomorphic to a subgroup of N_2: The same argument as the previous step, reversing the roles of N_1 and N_2.
  8. N_1 and N_2 are isomorphic, and equal each other's centralizers in \operatorname{Sym}(S): This follows from the previous two steps, and the fact that they are both finite groups. Note that this settles parts (1) and (2) of what we need to prove.
  9. We now construct the involution in \operatorname{Sym}(S) that conjugates N_1 to N_2: Note that we can think of S as a set identified with both N_1 and N_2, where N_1 acts by left multiplication and N_2 acts via right multiplication by the inverse element. Consider now a bijection from S to itself that sends each element of S to its inverse (when identified with either N_1 or N_2). This map conjugates left multiplication by an element of N_1 with right multiplication by the inverse of its corresponding element in N_2. This settles part (3) of what we need to prove.
  10. The action is primitive: We prove this by showing that if the action is not primitive, we can construct strictly smaller nontrivial normal subgroups of G contained in N_1.