Prove that for all fixed such that k isn't 0 the map defined by is an automorphism of .

So I need to show it is an isomorphism. I am having a problem with the homomorphism part though.

Let . but .

What do I need to do?

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- October 7th 2011, 02:01 PMdwsmithQ\to Q by q\to kq
Prove that for all fixed such that k isn't 0 the map defined by is an automorphism of .

So I need to show it is an isomorphism. I am having a problem with the homomorphism part though.

Let . but .

What do I need to do? - October 7th 2011, 02:10 PMSwlabrRe: Q\to Q by q\to kq
- October 7th 2011, 02:12 PMdwsmithRe: Q\to Q by q\to kq
- October 7th 2011, 02:14 PMSwlabrRe: Q\to Q by q\to kq
- October 7th 2011, 11:48 PMzoekRe: Q\to Q by q\to kq
- October 8th 2011, 12:00 AMDevenoRe: Q\to Q by q\to kq
in dwsmith's defense, it is bad form to speak of "the group Q". a group is not merely a set, but a set and a binary operation. failure to specify the intended binary operation requires an unwarranted amount of mysticism on the student's part.