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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 at 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> 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>. | ||
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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\}\,\,=\,\,\{0,5,10,15,20,25,30,35,40,45\}</math>. | ||
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| + | 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)\,\,=\,\,\{1,6,11,16,21,26,31,36,41,46\}</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) | ||
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| + | A simple check to make sure that the you find the Ker and Img is to see if the index of the ker times the index of the image is equal to the size of the the set initially. This works for any mapping that maps the same number of points to each element of the image, which this mapping does. | ||
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| + | -Allen | ||
Latest revision as of 15:42, 5 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)
A simple check to make sure that the you find the Ker and Img is to see if the index of the ker times the index of the image is equal to the size of the the set initially. This works for any mapping that maps the same number of points to each element of the image, which this mapping does.
-Allen
