Suppose $\phi:Z_{50}\rightarrow Z_{15}$ is a group homomorphism with $\phi(7)=6$.
Determine $\phi(x)$
My attempt: Since $7$ is a generator for $Z_{50}$,we can find other images as $$φ(7+7) = 6+6 = 12, φ(7 + 7 + 7) = 6 + 6 + 6 = 3, φ(7 + 7 + 7 + 7 + 7) = 9,...$$ etc
In the solution,its given as $$φ(x) = φ(x · 1) = xφ(1)$$,I know how to calculate $\phi(1)$,but I didn't undertsand how $\phi(x)=x$ in the equation
Note that $\mathbb Z_{50}$ and $\mathbb Z_{15}$ are groups under addition. Therefore, any homomorphism must satisfy $\phi(x) +_{15} \phi(y) = \phi(x+_{50}y)$ where $+_{k}$ indicates addition mod $k$. Now, If it is given that $\phi(7)=6$, then because $Z_n$ is cyclic, $\phi$ is determined at each value.
Furthermore, by the hint above, you should use $\phi(7)$ to find $\phi(1)$, because once you do this, $\phi(x) = x \times \phi(1)$ by the homomorphism property, where multiplication is mod $15$.
Now, what is $\phi(1)$? It's quite easy : we need to find the number $z$ such that $7z$ and $1$ are equivalent mod $50$ i.e. $7x-1$ is a multiple of $50$, because then $\phi(7z) = \phi(1)$. What we can find with a little help is that $z=43$ satisfies this relation. Therefore, $43 \times \phi(7) = \phi(43\times 7) = \phi(1)$. Therefore, the left hand side is $258$, which is $3 \mod 15$. Hence $\phi(1)=3$ and $\phi(x) = 3x$ for all $x$.
How do we know this works? Well, $\phi(7) = 3 \times 7 = 21 \equiv 6 \mod 15$, so $\phi(7) = 6$, and now by the fact that the value of $\phi$ and $3x$ are the same at a generator, they are the same everywhere.