Note that the question is asked in exercises for rings yet the question just mentions homomorphisms and doesn't specify group or ring. I am guessing this is because it is both a group and ring homomorphism?
I was trying to follow (https://math.stackexchange.com/q/1869580)'s answer but it was kind of hard because I didn't understand how $\left \langle 1\;\; (mod\;\;4) \right \rangle= \Bbb{Z}/4\Bbb{Z}$. Isn't it $\left \langle 1+4\Bbb{Z} \right \rangle =\Bbb{Z}/4\Bbb{Z}$?
By using Lagrange's theorem and properties of homomorphisms, I found that |$\phi(1 + 6\Bbb{Z})|=\left \{ 1,3 \right \}$. The problem for me is finding the set $\phi(1+6\Bbb{Z})$ since I don't know how to determine individual orders of the elements in $\Bbb{Z}/15\Bbb{Z}=\left \{15\Bbb{Z},…,14+15\Bbb{Z} \right \}$.
Am I doing anything wrong and is my notation awkward?
Best Answer
Let's look at group homomorphisms first.
If $\def\Z{\mathbb{Z}}f\colon\Z/6\Z\to\Z/15\Z$ is given, then it is determined by $f(1+6\Z)=a+15\Z$ and it must be $$ 6(a+15\Z)=0+15\Z $$ that is, $6a\equiv0\pmod{15}$ that immediately becomes $a\equiv0\pmod{5}$. So the only possibilities for $a$ are $0$, $5$ and $10$.
Each of these possibilities gives indeed a homomorphism, as you can check.
Now, a ring homomorphism is also a group homomorphism. If your definition of ring homomorphism requires preserving the identity, none of those satisfies it.
Suppose we have $f(1+6\Z)=5+15\Z$. Then $f(2+6\Z)=10+15\Z$ and $f(4+6\Z)=5+15\Z$. What's $(10+15\Z)^2$?