Which are the homomorphisms between the two groups $G=Z/(3)$ and $H=Z/(5)$?

Question

Given two groups $G$ and $H$, there is at least one homomorphism between them, namely the trivial homomorphism: $f:G\rightarrow H, \ f(x)=e_{H}, \forall x\in G.$ To specify a homomorphism $f:G\rightarrow H$ you just have to specify an element $h\in H$ to be the image of $e_{G}$ and any choice of $h$ gives you a well defined homomorphism. So there are |H| many homomorphisms $G\rightarrow H.$ But i just read in a book that for $G=Z/(3)$ and $H=Z/(5)$ there can be defined only the trivial homomorphism, which i dont understand. It is also not quite clear, what $Z/(3)$ means. I presume though that it means $Z/3Z.$

Can somebody comment on why between these two groups there can be defined only the trivial homomorphism and no other one ? Thanks for any comment.

Answer 1

Let's recall a few results. Let $\phi : G \to H$ be a group homomorphism, where $G, H$ are finite. Then:

• $\text{ker}(\phi)$ is a subgroup of $G$. So $|\text{ker}(\phi)|$ divides $|G|$.
• $\text{Im}(\phi)$ is a subgroup of $H$, so $|\text{Im}(\phi)|$ divides $|H|$
• $G/\text{ker}(\phi) \cong \text{Im}(\phi)$. Thus, $|G| = |\text{ker}(\phi)| * |\text{Im}(\phi)|$. In particular, $|\text{Im}(\phi)|$ divides $|G|$.

Now here we have $G = \mathbb{Z}_{3}$ and $H = \mathbb{Z}_{5}$. So $|\text{Im}(\phi)| \in \{1, 5\}$ by the second bullet point, and $|\text{Im}(\phi)| \in \{ 1, 3\}$ by the third bullet point. Thus, $|\text{Im}(\phi)| = 1$, which implies that $\phi$ is trivial.

Answer 2

Let $\phi:G\to H$ be a homomorphism. We must send $0$ to $0$. So all that remains is to specify where to send $1$ and $2$. Let $\phi(1)=a\in H$. Since $3=0$ in $G$, $\phi(3)=\phi(0)$, so $\phi(1)+\phi(1)+\phi(1)=3a=0_H$. But this can only happen if $a=0_H$. It follows that $\phi(2)=\phi(1)+\phi(1)=0_H$.

Answer 3

First of all, yes, $\mathbb{Z}/(n)=\mathbb{Z}/n\mathbb{Z}$ and, yes, there is no non-trivial group homomorphism from $\mathbb{Z}/3\mathbb{Z}$ into $\mathbb{Z}/5\mathbb{Z}$.

Now, why do you say that “To specify a homomorphism $f\colon G\longrightarrow H$ you just have to specify an element $h\in H$ to be the image of $e_G$ and any choice of $h$ gives you a well defined homomorphism”? The image of $e_G$ can only be $e_H$.

Answer 4

We know that, if $A$ is a commutative ring, $I$ an ideal of $A$ and $M$ is an $A$-module, then $$\DeclareMathOperator{\Hom}{Hom}\Hom(A/I,M)\simeq 0:_M I.$$ Apply this to the abelian group (= $\mathbf Z$-modules) $\mathbf Z/5\mathbf Z$ and the ideal $3\mathbf Z$: $$\Hom(\mathbf Z/3\mathbf Z,\mathbf Z/5\mathbf Z)\simeq 0:_{\mathbf Z/5\mathbf Z}3\mathbf Z. $$ Now, in $\mathbf Z/5\mathbf Z$, multiplication by $3$ is invertible since $3$ and $5$ are coprime, so the annihilator of $3\mathbf Z$ in $\mathbf Z/5\mathbf Z$ is $0$, and $$\Hom(\mathbf Z/3\mathbf Z,\mathbf Z/5\mathbf Z)=\{0\}.$$