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Subgroup/Normal Subgroup/Automorphism QuestionsIn this problem, assume (important!) that G is an Abelian. Set H = {g in G : g^5 = e}. (Warning: Expressions such as x^1/5 are not well defined. Do not use them!) (a) Show H is a subgroup of G. (b) Show H is a normal subgroup of G. (c) Assume further that G is finite and that H = {e}. Show that the map phi : G --> G given by phi(g) = g^5 is an automorphism. (Definition: An automorphism is an isomorphism from a group to itself.) |

Re: Subgroup/Normal Subgroup/Automorphism QuestionsQuote:
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If not, show it is injective: show that a?b -> phi(a)?phi(b). An injective function from A->A must be bijective for any finite set A. Cheers. |

Re: Subgroup/Normal Subgroup/Automorphism QuestionsQuote:
If not, show it is injective: show that a?b -> phi(a)?phi(b). An injective function from A->A must be bijective for any finite set A. Cheers.[/quote:16v4wr01] concerning part c... we've learned about homomorphisms. isn't the kernel of phi = e or something like that? i don't really know how to even begin showing part c. |

Re: Subgroup/Normal Subgroup/Automorphism QuestionsThe kernel is the set of elements which map to e. You should have learned that a homomorphism is n-1: every element in the range has the same number (n) of elements map to it. If you can show that only one element maps to e, then you have a 1-1, homomorphism. And since the sets are finite and the same size, this is a bijective homomorphism-- an isomorphism. |

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