Homomorphism From a Finite to an Infinite Group

Question

I know that any homomorphism will take the identity of the first group to the identity of the second; So assuming that $f$ is a function that takes $Z_{50} \to Z$ by any homomorphism of groups, I'm curious as to how to show that $f(1) = 0$ ?

I am struggling to see the connections from the finite group of $Z_{50}$ to the infinite group of $Z$ and how the function mapping would look..

I'm kinda a newbie so any and all help is appreciated

Answer 1

Since the domain is finite, so the image of the domain under $f$ will be a finite subgroup of $\Bbb{Z}$. But the only finite subgroup of $(\Bbb{Z},+)$ is $\{0\}$. Thus only the zero homomorphism is possible.

If you need more explanation for this, then please feel free to ask.

Answer 2

You want that $50\cdot f(1)=f(50)=f(0)=0$ so $f(1)$ is an element of $\Bbb Z$ that multiplied for 50 gives 0... $f(1)=0$!

Answer 3

You know that $f(0)=0$ because it is a homomorphism. You know too that $0\equiv_{50}50$. So f(50)=0. Since f is a homomorphism, you know that $f(1+1+1...+1)$ (fifty times) is equal to $50f(1)$, therefore you have that $50f(1)=0$. Like $\mathbb{Z}$ is an integral domain, you have $f(1)=0$