Finding a homormorphism form $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/6\mathbb{Z}$

Question

I've got the following question: find a homormorphism form $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/6\mathbb{Z}$.

My work so far: The group $\mathbb{Z}/N\mathbb{Z}$ has a generator of 1, so: If we have a homomorphism from $\mathbb{Z}/4\mathbb{Z}$ to $\mathbb{Z}/6\mathbb{Z}$ which sends $1 \in \mathbb{Z}/4\mathbb{Z}$ to an $m\in \mathbb{Z}/6\mathbb{Z}$ with $m \cdot 4=\bar{0}$ , does this mean that if $4m=\bar{0}$ in $\mathbb{Z}/6\mathbb{Z}$ implies $2m=\bar{0}$ and thus $m=0 \mod 2$? How should i finish this? Does this mean that there are three homomorphisms, namely you can send 1 t0 0, 2 and 4? I have read another question regarding the homomorphisms between $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$, but when I tried to apply it, I got stuck

Answer 1

In fact there are two homomorphisms. The image of $\mathbb{Z}_4$ is a subgroup of $\mathbb{Z}_6$, so it has to be of order $2$ or $1$.

As generators have to be sent to generators we have 2 homomorphism. The first one is given by the map $1 \to 3$ and the later is the trivial one given by $1 \to 0$.

Answer 2

I can finish this defining $f\colon\mathbb{Z}\longrightarrow\mathbb Z$ by $f(n)=3n$. Then $f(4\mathbb{Z})=12\mathbb{Z}\subset6\mathbb{Z}$. Therefore, $f$ induces a non-trivial homomorphism $f^\star\colon\mathbb{Z}/4\mathbb{Z}\longrightarrow\mathbb{Z}/4\mathbb{Z}$:

• $f^\star(\bar0)=\bar0$;
• $f^\star(\bar1)=\bar3$;
• $f^\star(\bar2)=\bar0$;
• $f^\star(\bar3)=\bar3$.

Answer 3

There are only two homomorphisms from $\mathbb{Z}/4\mathbb{Z}$ to $ \mathbb{Z}/6\mathbb{Z}$ and they are:

1) The trivial homomorphism taking everything to $\bar 0 $

2) The homomorphism $\bar 0 \to \bar 0$ and $\bar 1 \to \bar 3$ and $\bar 2 \to \bar 0$ and $\bar 3 \to \bar 3$

The reason for that is if $ x\equiv y \pmod 4$ then $ 3x \equiv 3y \pmod 6 $