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 \}.

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