Problem 5.4.18 (Normal subgroups of $K_1 \times K_2 \times \cdots K_n$, where the $K_i$ are non abelian simple groups)

Proof. Suppose that $N⊴G$, and let $x=(a_1,\dots ,a_n)$ be an element of $N$ such that $a_i\ne 1$.

Assume for the sake of contradiction that for all $g\in K_i$, $g{a}_i=a_{i}g$. Then $g{a}_i{g}^{-1}=a_i$, thus $⟨a_i⟩⊴K_i$. But $K_i$ is simple, and $⟨a_i⟩\ne \{1\}$, therefore $K_i=⟨a_i⟩$, so $K_i$ is abelian, which contradicts the hypothesis. This contradiction shows that

Let ${\tilde{K}}_i$ denote the set of elements $(1,\dots ,a,\dots ,1)$, where $a\in K_i$, so that $\widetilde{K_i}\simeq K_i$ is simple ( ${\tilde{K}}_i$ is identified with $K_i$ in the statement).

Put ${\tilde{g}}_i=(1,\dots ,g_i,\dots ,1)\in \widetilde{K_i}$. Since $N⊴G$, and $x\in N$,

and

where $g_i^{-1}{a}_i^{-1}{g}_i{a}_i\ne 1$, so

Therefore ${\tilde{K}}_i\cap N\ne \{1\}$, and ${\tilde{K}}_i\cap N⊴{\tilde{K}}_i$ (since $N⊴G$). But $\widetilde{K_i}$ is simple, thus ${\tilde{K}}_i\cap N={\tilde{K}}_i$, and so

If $I=\{i_1,\dots ,i_k\}\subseteq [[1,n]]$ is the set of $i$ such that there is some $x=(a_1,\dots ,a_n)\in N$ such that $a_i\ne 1$, then ${\tilde{K}}_i\le N$, and $N\cap {\tilde{K}}_j=\{1\}$ if $j\notin I$, therefore

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