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--[[User:Robertsr|Robertsr]] 10:24, 1 October 2008 (UTC) | --[[User:Robertsr|Robertsr]] 10:24, 1 October 2008 (UTC) | ||
| + | |||
| + | ---- | ||
| + | |||
| + | For the remaining two elements: | ||
| + | |||
| + | <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> | ||
| + | |||
| + | :--[[User:Narupley|Nick Rupley]] 01:34, 2 October 2008 (UTC) | ||
Revision as of 20:34, 1 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)
