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:--[[User:Nmisner|Neely Misner]] 12:42, 2 October 2008 (UTC) | :--[[User:Nmisner|Neely Misner]] 12:42, 2 October 2008 (UTC) | ||
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| + | I could be wrong, but I don't think there is actually a closed form for this particular homomorphism. At least, I wasn't able to find one. We have, however, completely defined <math>\scriptstyle\phi</math> for all of its domain elements, so that suffices as a solution to the exercise. | ||
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| + | :--[[User:Narupley|Nick Rupley]] 12:40, 2 October 2008 (UTC) | ||
Revision as of 08:40, 2 October 2008
We know that phi(7) = 7 and thus we can see that phi7 * 7) = phi(7) * phi(7) by the morphism law which is 49 -30 which equals 19.
Same applies with phi(7^3) it equals 13.
How do we get the elements 23 and 29 to finish the homomorphism?
--Robertsr 10:24, 1 October 2008 (UTC)
For the remaining two elements:
$ \scriptstyle\phi(13*11)\,\,=\,\,\phi(143)\,\,=\,\,\phi(23)\,\,=\,\,\phi(13)\phi(11)\,\,=\,\,13*1\,mod\,30\,\,=\,\,13 $
$ \scriptstyle\phi(19*11)\,\,=\,\,\phi(209)\,\,=\,\,\phi(29)\,\,=\,\,\phi(19)\phi(11)\,\,=\,\,19*1\,mod\,30\,\,=\,\,19 $
- --Nick Rupley 01:34, 2 October 2008 (UTC)
I don't quite understand this problem. I get how to map things from U(30) to U(30) but what is the actual morphism that we are trying to find?
- --Neely Misner 12:42, 2 October 2008 (UTC)
I could be wrong, but I don't think there is actually a closed form for this particular homomorphism. At least, I wasn't able to find one. We have, however, completely defined $ \scriptstyle\phi $ for all of its domain elements, so that suffices as a solution to the exercise.
- --Nick Rupley 12:40, 2 October 2008 (UTC)
