Given the function $f:\Bbb Z_{40} \rightarrow \Bbb Z_{60}$, how many of them are there such that $f([0]_{40})=[0]_{60}$ and $f([1]_{40})=[1]_{60}$? How many of them are homomorphism of additive groups?
I have no idea how to answer to these questions. Any help? For the first question, is it a good way set up a system to know when $x\equiv(0 \mod 60) $ and $x\equiv(1 \mod 60)$?
For a homomorphism you have $f(a^n)=f(a)^n$ or in additive notation $f(na)=nf(a)$ so we have that $f(n)=nf(1)$, so a homomorphism is defined by $f(1)$ you also need $$f(0)=f(39+1)=f(39)+f(1)=40f(1)=0$$ When is $40f(1)=0$? How many choices are there for $f(1)$?
For the first question you only have $f(0)=0$ and $f(1)=1$ then for other $38$ elements of $Z_{40}$ you can map each to any of the $60$ elements of $Z_{60}$ hence it's $60^{38}$