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| − | The function <math>\scriptstyle\phi:Z_{50}\to Z_{15}</math> | + | The function <math>\scriptstyle\phi:Z_{50}\to Z_{15}</math> maps all of <math>\scriptstyle Z_{50}</math> to <math>\scriptstyle 3Z_5\subset Z_{15}</math>, such that <math>\scriptstyle im(\phi)\,\,=\,\,\{0,3,6,9,12\}</math>. |
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>: | 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\}\,\,=\,\,\{0,5,10,15\}</math>. | + | <math>\scriptstyle ker(\phi)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=0\}\,\,=\,\,\{0,5,10,15,20,25,30,35,40,45\}</math>. |
We know that <math>\scriptstyle\phi(1)\,\,=\,\,3*1\,mod\,15\,\,=\,\,3</math>. 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>: | We know that <math>\scriptstyle\phi(1)\,\,=\,\,3*1\,mod\,15\,\,=\,\,3</math>. 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\}\,\,=\,\,1+ker(\phi)\,\,=\,\,\{1,6,11,16\}</math>. | + | <math>\scriptstyle\phi^{-1}(3)\,\,=\,\,\{x\in Z_{50}\,|\,\phi(x)=3\}\,\,=\,\,1+ker(\phi)\,\,=\,\,\{1,6,11,16,21,26,31,36,41,46\}</math>. |
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| + | *EDIT: I made some mistakes in my initial solution, the function <math>\scriptstyle\phi</math> is not surjective to <math>\scriptstyle Z_{15}</math>, thus the kernel (and subsequently <math>\scriptstyle\phi^{-1}(3)</math>) are larger than I had originally stated. | ||
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:--[[User:Narupley|Nick Rupley]] 22:29, 1 October 2008 (UTC) | :--[[User:Narupley|Nick Rupley]] 22:29, 1 October 2008 (UTC) | ||
Revision as of 10:21, 2 October 2008
Problem: Suppose \phi: Z_50 -> Z_15 is a group homomorphism with phi(7) = 6
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 at 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} $ maps all of $ \scriptstyle Z_{50} $ to $ \scriptstyle 3Z_5\subset Z_{15} $, such that $ \scriptstyle im(\phi)\,\,=\,\,\{0,3,6,9,12\} $.
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\}\,\,=\,\,\{0,5,10,15,20,25,30,35,40,45\} $.
We know that $ \scriptstyle\phi(1)\,\,=\,\,3*1\,mod\,15\,\,=\,\,3 $. 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\}\,\,=\,\,1+ker(\phi)\,\,=\,\,\{1,6,11,16,21,26,31,36,41,46\} $.
- EDIT: I made some mistakes in my initial solution, the function $ \scriptstyle\phi $ is not surjective to $ \scriptstyle Z_{15} $, thus the kernel (and subsequently $ \scriptstyle\phi^{-1}(3) $) are larger than I had originally stated.
- --Nick Rupley 22:29, 1 October 2008 (UTC)
