Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 3 / c

A finite minimal normal subgroup is a characteristically simple group, and has the direct-product structure of a finite minimal normal subgroupwith isomorphic simple groups $T_i$. For completeness, choose a minimal nontrivial normal subgroup $T$ of $N$. Its distinct $G$-conjugates are minimal normal subgroups of $N$, so they centralize one another. Their product is normal in $G$, hence equals $N$. If $T$ is abelian the product is abelian and minimal primary characteristic subgroups make $N$ elementary abelian. Otherwise each $T$ has trivial center: its center is a characteristic subgroup of $T$, hence normal in $N$, and minimality would make $T$ abelian. Each new conjugate then meets the product of previous conjugates trivially, because it centralizes that product and is centerless. The product is direct. Every normal subgroup of a direct factor is normal in $N$, so minimality of that factor makes it simple.

Assume now that $N$ acts primitively and nonregularly. Any nontrivial normal subgroup of this faithful primitive action is transitive. Thus each $T_i$ is transitive. If $k\ge 2$, the transitive commuting groups $T_1$ and $T_2$ are regular: an element of one fixing $\alpha$ commutes with the other and hence fixes its entire orbit. But the centralizer of a regular group has order $n$, as its elements are determined by the image of $\alpha$. Therefore all remaining factors lie in the order-n centralizer of $T_1$, and the transitive group $T_2$ already fills it. There can be only two factors, with $|N|=n^2$. Now $|N_{\alpha }|=n$, whereas its common nontrivial subdegree $r$ divides both $n$ and $n-1$. Hence $r=1$, making $N_{\alpha }$ trivial, a contradiction. Thus $k=1$, and $N$ is simple. An abelian transitive group is regular, so $N$ is nonabelian.

Finally $C_G(N)$ is normal in the primitive group $G$. If nontrivial it would be transitive, and commuting with transitive $N$ would force $N$ regular. Thus $C_G(N)=1$, and conjugation embeds $G$ in $\mathrm{Aut}(N)$. The centerless simple group $N$ is identified with its inner automorphism group. Therefore