# 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.

## 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 $N_1$.