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Let S be a simple group and set $G = S \times ... \times S$. I believe that G has no proper characteristic subgroups but I have no idea about how to prove this. Any help or counterexamples would be appreciated.

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up vote 4 down vote accepted

If S is non-abelian then first classify the normal subgroups G. This is very easy and follows from the coordinate-wise definition of conjugation. Now observe that clearly none of these normal subgroups is characteristic by exhibiting an automorphism moving them.

If S is abelian, then this is just linear algebra: every subgroup of the same order is conjugate under an automorphism.

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