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--[[User:Robertsr|Robertsr]] 10:24, 1 October 2008 (UTC) | --[[User:Robertsr|Robertsr]] 10:24, 1 October 2008 (UTC) | ||
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| + | ---- | ||
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| + | For the remaining two elements: | ||
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| + | <math>\scriptstyle\phi(13*11)\,\,=\,\,\phi(143)\,\,=\,\,\phi(23)\,\,=\,\,\phi(13)\phi(11)\,\,=\,\,13*1\,mod\,30\,\,=\,\,13</math><br> | ||
| + | <math>\scriptstyle\phi(19*11)\,\,=\,\,\phi(209)\,\,=\,\,\phi(29)\,\,=\,\,\phi(19)\phi(11)\,\,=\,\,19*1\,mod\,30\,\,=\,\,19</math> | ||
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| + | :--[[User:Narupley|Nick Rupley]] 01:34, 2 October 2008 (UTC) | ||
| + | |||
| + | ---- | ||
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| + | 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? | ||
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| + | :--[[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) | ||
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| + | ---- | ||
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| + | There was no closed form that I found either. But, there is no need for a closed form if you can map all the elements for U(30). I solved this problem by noting that phi^(-1)(x) = x*Ker(phi). This allows you to find which values map to each member of U(30), because the kernel is given. | ||
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| + | :--[[User:Nclaus|Nclaus]] 18:15, 4 October 2008 (UTC) | ||
Latest revision as of 13:15, 4 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)
There was no closed form that I found either. But, there is no need for a closed form if you can map all the elements for U(30). I solved this problem by noting that phi^(-1)(x) = x*Ker(phi). This allows you to find which values map to each member of U(30), because the kernel is given.
- --Nclaus 18:15, 4 October 2008 (UTC)
