1. ## group homomorphism challenges

Here is the question:
Suppose that ϕ: Z_50 to Z_15 is a group homomorphism with ϕ(7) = 6
a)determine ϕ(x)
b) determine the image of ϕ
c) determine the kernel of ϕ
d) determine ϕ^(-1)(3). That is, determine the set of all elements that map to 3.

--- a) doesn't ϕ(x) = (6/7)x
b) ok, so the image of ϕ is something that i am confused on, we didn't really talk about in class and was barely covered in my book. My professor told me that the image of ϕ is denoted phi(G) and is the set of all elements of G-bar that are 'hit' by ϕ. so does this mean that the image of ϕ is any element in Z_15 such that that element times 6/7 gets me an element in Z_50?
c)I understand the kernel. It is any element in Z_50 times 6/7 that gets me the element, that is 0 in Z_15 (an element of 15). but there are not many of these, only {0, 35} i think this is right, but i am a little skeptical about ϕ(x)=6/7
d)so this is similar to the kernel, but instead of mapping to the identity, 0, it maps to three? if that is the case then i know that 21 is an example, but otherwise i am having a difficult time. thank you for any help.

2. ## Re: group homomorphism challenges

a) try not to use "fractions" for elements of Z15. what you want to do is notice that 7 generates Z50, since gcd(7,50) = 1. now, usually we think of elements of Z50 as being "multiples" of 1, that is: x = 1+1+...+1 (x times). note that 7^2 = 49 = -1 (mod 50), so 7^4 = 1 (mod 50). this means "1/7" is 7^3 (mod 50), which is 43.

so x = 1x = (43)(7)x = 43x(7) (mod 50). since φ is a homomorphism, φ(43x(7)) = 43xφ(7) = 43x(6) (mod 15).

but mod 15, 43x(6) = 13x(6) = 78x = 3x (mod 15).

3. ## Re: group homomorphism challenges

Ok this helps. But the sentence " this means "1/7" is 7^3 (mod 50), which is 43." is really confusing. I see why 7^3 is 43 but where do you get this 1/7 from? That is what is confusing.

4. ## Re: group homomorphism challenges

well, see, that's why fractions are confusing.

what i really mean is $\displaystyle 7^{-1}$ under multiplication mod 50.

5. ## Re: group homomorphism challenges

Thank you i understand it now. My last question is i found from spmeone that the answer to b is 0,3,6,9. But why? This i just dont understand. Everything else i get.

6. ## Re: group homomorphism challenges

well, we know that φ(1) = 3, right?

so φ(2) = φ(1+1) = φ(1) + φ(1) = 3 + 3 = 6,
φ(3) = φ(1+1+1) = φ(1) + φ(1) + φ(1) = 3 + 3 + 3 = 9
φ(4) = φ(1+1+1+1) = φ(1) + φ(1) + φ(1) + φ(1) = 3 + 3 + 3 + 3 = 12
φ(5) = φ(1+1+1+1+1) = φ(1) + φ(1) + φ(1) + φ(1) + φ(1) = 3 + 3 + 3 + 3 + 3 = 15 = 0.

now φ(k + 5n) = φ(k) + φ(5n) = φ(k) + nφ(5) = φ(k) + n(0) = φ(k), so we're not going to get any more new values for φ(k).