(New page: We know that if <math>\scriptstyle\phi(g)\,\,=\,\,g^{\prime}</math>, then <math>\scriptstyle\phi^{-1}(g^{\prime})\,\,=\,\,\{x\in G\,|\,\phi(x)\,=\,g^{\prime}\}\,\,=\,\,g\,ker(\phi)</math>....)
 
 
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We know that if <math>\scriptstyle\phi(g)\,\,=\,\,g^{\prime}</math>, then <math>\scriptstyle\phi^{-1}(g^{\prime})\,\,=\,\,\{x\in G\,|\,\phi(x)\,=\,g^{\prime}\}\,\,=\,\,g\,ker(\phi)</math>.
 
We know that if <math>\scriptstyle\phi(g)\,\,=\,\,g^{\prime}</math>, then <math>\scriptstyle\phi^{-1}(g^{\prime})\,\,=\,\,\{x\in G\,|\,\phi(x)\,=\,g^{\prime}\}\,\,=\,\,g\,ker(\phi)</math>.
  
So, if <math>\scriptstyle\phi\,:\,U(30)\to U(30)\,=\,\{1,2,3,5,7,11,13,17,19,23,29\}</math>, <math>\scriptstyle ker(\phi)\,\,=\,\,\{1,11\}</math>, and <math>\scriptstyle\phi(7)\,\,=\,\,7</math>, then:
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So, if <math>\scriptstyle\phi\,:\,U(30)\to U(30)\,=\,\{1,7,11,13,17,19,23,29\}</math>, <math>\scriptstyle ker(\phi)\,\,=\,\,\{1,11\}</math>, and <math>\scriptstyle\phi(7)\,\,=\,\,7</math>, then:
  
 
<math>\scriptstyle\phi^{-1}(7)\,\,=\,\,7\,ker(\phi)\,\,=\,\,\{7*1\,mod\,30,7*11\,mod\,30\}\,\,=\,\,\{7,17\}</math>.
 
<math>\scriptstyle\phi^{-1}(7)\,\,=\,\,7\,ker(\phi)\,\,=\,\,\{7*1\,mod\,30,7*11\,mod\,30\}\,\,=\,\,\{7,17\}</math>.
  
 
:--[[User:Narupley|Nick Rupley]] 23:58, 1 October 2008 (UTC)
 
:--[[User:Narupley|Nick Rupley]] 23:58, 1 October 2008 (UTC)

Latest revision as of 21:11, 1 October 2008

We know that if $ \scriptstyle\phi(g)\,\,=\,\,g^{\prime} $, then $ \scriptstyle\phi^{-1}(g^{\prime})\,\,=\,\,\{x\in G\,|\,\phi(x)\,=\,g^{\prime}\}\,\,=\,\,g\,ker(\phi) $.

So, if $ \scriptstyle\phi\,:\,U(30)\to U(30)\,=\,\{1,7,11,13,17,19,23,29\} $, $ \scriptstyle ker(\phi)\,\,=\,\,\{1,11\} $, and $ \scriptstyle\phi(7)\,\,=\,\,7 $, then:

$ \scriptstyle\phi^{-1}(7)\,\,=\,\,7\,ker(\phi)\,\,=\,\,\{7*1\,mod\,30,7*11\,mod\,30\}\,\,=\,\,\{7,17\} $.

--Nick Rupley 23:58, 1 October 2008 (UTC)

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