For part a) I got phi(x)=3x. For the kernel (part c)) I got {0,5,10,15,20,25,30,35,40,45}. I wasn't really sure how to do the image and the inverse though. Does anyone have any ideas? I thought that the inverse might be 3*ker(phi) by theorem 10.6.1, but I wasnt sure. --Clwarner 14:44, 18 February 2009 (UTC)
Yea I got the same thing in part a by showing:
7^2 = 49 = -1 mod 50
φ(-1) = φ(49) = φ(7*7) = 7*φ(7) = 7*6 = 42 = 12 mod 15
φ(1) = -φ(-1) = -12 = 3 mod 15
Therefore, φ(x) = 3x. The answer checks out, but I don't know if this is the right way to get it.
You can get the inverse by looking at all the possible x values and then determining which of these values of x give us φ(x) = 3. We know x = {0,1,2,...,49}. To find which values of x give us 3*x mod 15 = 3, we could iterate all of the 3*x values as {0,3,6,9,12,...,147} and then determining which ones are equivalent to 3 in Z_15. It's easier however to use property 6 of Theorem 10.1 in the book which says that φ^-1(g') = gKer(φ). This means all we need is one value x such that 3x mod 15 = 3 and then add each value of Kerφ to that x to determine the inverse set. The simplest value for x is 1. Since our kernel is just the multiples of 5 in [0,45] we just get the inverse as {1+0, 1+5, 1+10,..., 1+45} = {1,6,11,16,...,46}.
The image I believe is just {0,3,6,9,12} since φ(x) = 3x. This is the multiples of 3 from 0 to 14.
--Jniederh 15:15, 18 February 2009 (UTC)
I think the image of φ is just the elements in Z mod 15Z that are being mapped to. So {0, 3, 6, 9, 12}.--Jcromer 15:58, 18 February 2009 (UTC)
