Homomorphism from $\mathbb{Z}_{120}$ to $\mathbb{Z}_{167}$

Question

Find all the morphisms from $\mathbb{Z}_{120}$ to $\mathbb{Z}_{167}$.

I've tried that:

Let $f : \mathbb{Z}_{120} \to \mathbb{Z}_{167}$ a homomorphism. Then let $x \in \mathbb{Z}_{120}$, then $x^{120} = e$. Because $f$ it's a homomorphism we have $f(xy) = f(x)f(y), \forall x,y \in \mathbb{Z}_{120}$.

But $f(x^{120}y) = f(x^{120})f(y) = f^{120}(x)f(y) = f(y)$, So $f^{120}(x) = e$. But $ord(\mathbb{Z}_{167}) = 167$ so then $f^{167}(x) = e$.

From that we get $f^{120}(x) = f^{167}(x) = e$. Hence $120$ and $167$ are coprime, we get $f(x) = e$ so $f(x) = e$ it's the only homomorphism.

Answer 1

The image of $f$ is a subgroup of $\mathbb{Z_{167}}$, hence the order of the image divides $167$ by Lagrange's theorem. Since $167$ is a prime number, we have $|\operatorname{Im}(f)|=1$ or $|\operatorname{Im}(f)|=167$. But obviously there are no surjective functions from a set with $120$ elements to a set with $167$ elements, so it must be $|\operatorname{Im}(f)|=1$.

Answer 2

You're right. There's no nontrivial homomorphism from $\Bbb Z_n$ to $\Bbb Z_m$ unless $\rm{gcd}(n,m)\gt1$.

This essentially follows from Lagrange, and the homomorphism property.

Answer 3

The number of homomorphisms from $\mathbb{Z}_m $ to $\mathbb{Z}_n $ is $\gcd(m,n) $. Clearly, $\gcd(120,167) = 1$, so, there is only a trivial homomorphism.