Exercise 14.3.16

Proof. If $G$ has a unique minimal normal subgroup $N$, then the socle $H$ is $N$, and by Proposition 14.3.10, $H=N$ is a product of finite simple groups.

Suppose now that $G$ contains two minimal normal subgroups $N_1\ne N_2$. Then $N_1\cap N_2$ is normal in $G$, and $N_1$ is a minimal normal subgroup, therefore $N_1\cap N_2=N_1$ or $N_1\cap N_2=\{e\}$, and using the minimality of $N_2$, $N_1\cap N_2=N_2$ or $N_1\cap N_2=\{e\}$. If $N_1\cap N_2\ne \{e\}$, then $N_1=N_1\cap N_2=N_2$, in contradiction with the assumption, thus $N_1\cap N_2=\{e\}$.

By Exercise 7, $N_1{N}_2$ is a normal subgroup of $G$, such that $(x_1,x_2)\mapsto x_1{x}_2$ is an isomorphism from $N_1\times N_2$ to $N_1{N}_2$.

Consider $\mathcal{}A$ the set of all subgroups of $G$ of the form $N_1\cdots N_k$ such that $N_1\cdots N_k$ is a normal subgroup of $G$, and such that the map $(x_1,\dots ,x_k)\mapsto x_1\cdots x_k$ is an isomorphism from $N_1\times \cdots \times N_k$ to $N_1\cdots N_k$.

Then pick $A=N_1\cdots N_k$ an element of $\mathcal{}A$ of maximal order. Reasoning by contradiction, suppose that $A\ne H$. Then there is some minimal normal subgroup $N$ such that $N\not\subset A$, otherwise the socle, which is the subgroup $H$ generated by the minimal normal subgroups of $G$ would be $A=N_1\cdots N_k$. Write $N_{k+1}=N$.

Then $N_1\cdots N_k\cap N_{k+1}\ne N_{k+1}$, thus the minimality of $N$ shows that $N_1\cdots N_k\cap N_{k+1}=\{e\}$. Then Exercise 7 shows that $(N_1\cdots N_k)N_{k+1}$ is normal, and the map $(x_1,\dots ,x_k)\mapsto x_1\cdots x_{k+1}$ is an isomorphism from $N_1\times \cdots \times N_{k+1}$ to $N_1\cdots N_{k+1}$.

Therefore $N_1\cdots N_{k+1}\in \mathcal{}A$, and since $|N_1\cdots N_{k+1}|>|N_1\cdots N_k|=|A|$, this contradicts the maximality of $A$.

This proves $H=N_1\cdots N_k$, so that the socle $H=N_1\cdots N_k\simeq N_1\times \cdots N_k$ is a direct product of minimal normal subgroups, and by Proposition 14.3.10, each of them is isomorphic to a direct product of simple group. We have proved that $H$ is isomorphic to a product of finite simple groups. □