## Characterize the normal subgroups of a direct product of nonabelian simple groups

Let $K_1$, …, $K_n$ be nonabelian simple groups and let $G = K_1 \times \cdots \times K_n$. Prove that every normal subgroup of $G$ is of the form $G_I$ for some subset $I \subseteq \{1,2,\ldots,n\}$ (where $G_I$ is as defined in this previous exercise). [Hint: If $N \leq G$ is normal and $x = (a_i) \in N$ with some $a_i \neq 1$, show that there exists $g_i \in G_i$ not commuting with $a_i$. Show then that $[(1,\ldots,1,g_i,1,\ldots,1),x] \in K_i \cap N$ and deduce that $K_i \leq N$.]

Suppose $N \leq G$ is normal and nontrivial, and let $x = (a_i) \in N$ where $a_k \neq 1$ for some $k$. Since $K_k$ is simple and nonabelian, there exists $g_k \in K_k$ such that $g_ka_k \neq a_kg_k$, as otherwise $Z(K_k)$ is nontrivial.

Let $y = (b_i)$, where $b_i = g_k$ if $i = k$ and 1 otherwise.

Consider $[y,x] = y^{-1}x^{-1}yx$; note that since $N$ is normal, $y^{-1}x^{-1}y \in N$, hence $[y,x] \in N$. Moreover, $[y,x]_i = g_k^{-1}a_k^{-1}g_ka_k$ if $i=k$ and 1 otherwise, so that $[y,x] \in G_{\{k\}}$. Thus $[y,x] \in G_{\{k\}} \cap N$.

Now $G_{\{k\}} \cap N$ is a nontrivial normal subgroup of $G_{\{k\}} \cong K_k$, which is simple. Thus $G_{\{k\}} \cap N = G_{\{k\}}$, and we have $G_{\{k\}} \leq N$.

Thus $N = G_I$, where $I = \{ i \ |\ a_i \neq 1\ \mathrm{where}\ x_i = a_i\ \mathrm{for\ some}\ x \in N \}$.