11. Let G be a group which has a subgroup of index 6. Prove that G has a normal subgroup whose index is a divisor of 720.
Solution: Suppose that H is a subgroup with index 6. Letting G act by multiplication on the left cosets of H produces a homomorphism from G into S_{6}. The order of the image must be a divisor of | S_{6} | = 720, and so the index of the kernel is a divisor of 720.
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12. Let G act on the subgroup H by conjugation, let S be the set of all conjugates of H, and let µ : G -> Sym (S) be the corresponding homomorphism. Show that ker(µ) is the intersection of the normalizers N(aHa^{ -1}) of all conjugates.
Solution: We see that x is in ker(µ) if and only if x(aHa^{ -1})x^{ -1} = aHa^{ -1} for all a in G.
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