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'''Problem: Suppose \phi: Z_50 -> Z_15 is a group homomorphism with phi(7) = 6''' | '''Problem: Suppose \phi: Z_50 -> Z_15 is a group homomorphism with phi(7) = 6''' | ||
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| + | --- | ||
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| + | Suppose that <math>\scriptstyle\phi:Z_{50}\to Z_{15}</math> is a group homomorphism with <math>\scriptstyle\phi(7) = 6</math>. <math>\scriptstyle Z_{15}</math> is a group under the operation addition modulo 15. So, one would naturally assume that the homomorphism <math>\scriptstyle\phi</math> maps <math>\scriptstyle Z_{50}</math> to <math>\scriptstyle Z_{15}</math> using modulo 15. Let's take a stab it, guessing that <math>\scriptstyle\phi</math> is some linear function of x under modulo 15: | ||
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| + | <math>\scriptstyle 15*c+6\,=\,7*d+e\,\,|\,c,d,e\,\in\,\mathbb{Z}^{+}\cup\{0\}</math> | ||
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| + | This one just happens to be readily apparent: 15 + 6 = 21, so for our purposes, d = 3 and e = 0. So, <math>\scriptstyle\phi(x)\,=\,3x\,mod\,15</math>. | ||
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| + | To check, note that <math>\scriptstyle\phi(7)\,\,=\,\,21\,mod\,15\,\,=\,\,6</math>. | ||
| + | Also, <math>\scriptstyle\phi(a+b)\,\,=\,\,(3a+3b)\,mod\,15\,\,=\,\,3a\,mod\,15\,+\,3b\,mod\,15\,\,=\,\,\phi(a)+\phi(b)</math>. | ||
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| + | The function <math>\scriptstyle\phi:Z_{50}\to Z_{15}</math> is surjective, such that <math>\scriptstyle im(\phi)\,=\,Z_{15}</math>. | ||
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| + | Here, the kernel is the set of all elements in <math>\scriptstyle Z_{50}</math> which map to the identity (which is of course zero) under <math>\scriptstyle\phi</math>: | ||
| + | <math>\scriptstyle ker(\phi)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=0\}\,\,=\,\,\{0,15,30,45\}</math>. | ||
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| + | Similar to the kernel, <math>\scriptstyle\phi^{-1}(3)</math> is the set of all elements in <math>\scriptstyle Z_{50}</math> which map to 3 under <math>\scriptstyle\phi</math>: | ||
| + | <math>\scriptstyle\phi^{-1}(3)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=3\}\,\,=\,\,3+ker(\phi)\,\,=\,\,\{3,18,33,48\}</math>. | ||
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| + | :--[[User:Narupley|Nick Rupley]] 22:29, 1 October 2008 (UTC) | ||
Revision as of 17:29, 1 October 2008
Problem: Suppose \phi: Z_50 -> Z_15 is a group homomorphism with phi(7) = 6
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Suppose that $ \scriptstyle\phi:Z_{50}\to Z_{15} $ is a group homomorphism with $ \scriptstyle\phi(7) = 6 $. $ \scriptstyle Z_{15} $ is a group under the operation addition modulo 15. So, one would naturally assume that the homomorphism $ \scriptstyle\phi $ maps $ \scriptstyle Z_{50} $ to $ \scriptstyle Z_{15} $ using modulo 15. Let's take a stab it, guessing that $ \scriptstyle\phi $ is some linear function of x under modulo 15:
$ \scriptstyle 15*c+6\,=\,7*d+e\,\,|\,c,d,e\,\in\,\mathbb{Z}^{+}\cup\{0\} $
This one just happens to be readily apparent: 15 + 6 = 21, so for our purposes, d = 3 and e = 0. So, $ \scriptstyle\phi(x)\,=\,3x\,mod\,15 $.
To check, note that $ \scriptstyle\phi(7)\,\,=\,\,21\,mod\,15\,\,=\,\,6 $. Also, $ \scriptstyle\phi(a+b)\,\,=\,\,(3a+3b)\,mod\,15\,\,=\,\,3a\,mod\,15\,+\,3b\,mod\,15\,\,=\,\,\phi(a)+\phi(b) $.
The function $ \scriptstyle\phi:Z_{50}\to Z_{15} $ is surjective, such that $ \scriptstyle im(\phi)\,=\,Z_{15} $.
Here, the kernel is the set of all elements in $ \scriptstyle Z_{50} $ which map to the identity (which is of course zero) under $ \scriptstyle\phi $:
$ \scriptstyle ker(\phi)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=0\}\,\,=\,\,\{0,15,30,45\} $.
Similar to the kernel, $ \scriptstyle\phi^{-1}(3) $ is the set of all elements in $ \scriptstyle Z_{50} $ which map to 3 under $ \scriptstyle\phi $:
$ \scriptstyle\phi^{-1}(3)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=3\}\,\,=\,\,3+ker(\phi)\,\,=\,\,\{3,18,33,48\} $.
- --Nick Rupley 22:29, 1 October 2008 (UTC)
