Plinth theorem

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

 * Abelian minimal normal subgroup and core-free maximal subgroup are permutable complements
 * Abelian permutable complement to core-free subgroup is self-centralizing
 * Primitive implies Fitting-free or elementary Abelian Fitting subgroup

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:
 * 6) * 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 $$N_1$$.
 * 7) * 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.
 * 8) $$N_2$$ 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.
 * 9) $$N_1$$ 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 $$N_2$$.
 * 10) $$N_1$$ 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.
 * 11) 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.
 * 12) 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 $$N_1$$.