Let , …, be nonabelian simple groups and let . Prove that every normal subgroup of is of the form for some subset (where is as defined in this previous exercise). [Hint: If is normal and with some , show that there exists not commuting with . Show then that and deduce that .]
Suppose is normal and nontrivial, and let where for some . Since is simple and nonabelian, there exists such that , as otherwise is nontrivial.
Let , where if and 1 otherwise.
Consider ; note that since is normal, , hence . Moreover, if and 1 otherwise, so that . Thus .
Now is a nontrivial normal subgroup of , which is simple. Thus , and we have .
Thus , where .