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} $ 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\}\,\,=\,\,\{0,5,10,15\} $.
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\} $.
- --Nick Rupley 22:29, 1 October 2008 (UTC)
