Question

describe all homomorphisms f: Z4 - Z6. describe their kernels and ranges.

Answer 1

Question 1.

Describe all homomorphisms $f:\mathbb{Z}_{4}\to\mathbb{Z}_{6}$. Describe their kernels and ranges.

Solution. Since $\mathbb{Z}_{4}$ is cyclic and $[1]_{4}$ generates $\mathbb{Z}_{4}$, any homomorphism on this group is fully determined by $f([1]_{4})$. Note that

$4f([1]_{4})=f(4[1]_{4})=f([0]_{4})=[0]_{6},$

so the order of $f([1]_{4})$ in $\mathbb{Z}_{6}$ should divide $4$. The order of $[0]_{6}$ is $1$, the order of $[1]_{6}$ is $6$, the order of $[2]_{6}$ is $3$, the order of $[3]_{6}$ is $2$, the order of $[4]_{6}$ is $3$ and the order of $[5]_{6}$ is $6$. Thus, we have two cases: either $f([1]_{4})=[0]_{6}$, or $f([1]_{4})=[3]_{6}$.

If $f([1]_{4})=[0]_{6}$, then for any $n=0,\ldots,3$:

$f([n]_{4})=f(n[1]_{4})=nf([1]_{4})=n[0]_{6}=[0]_{6},$

so $f$ is the identity homomorphism. Then $\ker f=\mathbb{Z}_{4}$ and $\operatorname{im}f=\{[0]_{6}\}$.

Now assume $f([1]_{4})=[3]_{6}$. Then for any $n\in\mathbb{Z}$:

\[ f([n]_{4})=n[3]_{6}=[3n]_{6}=\begin{cases}[3]_{6},&n\text{ is odd},\\

[0]_{6},&n\text{ is even}.\end{cases} \] (1)

Check the correctness. Suppose $[m]_{4}=[n]_{4}$, i. e. $m-n=4k$ for some $k\in\mathbb{Z}$. Then $m$ is even if and only if $n$ is even. So, $f([m]_{4})=f([n]_{4})$. Thus, the equality (1) is correct, so it defines a homomorphism $f:\mathbb{Z}_{4}\to\mathbb{Z}_{6}$, because

$f([m]_{4}+[n]_{4})=f([m+n]_{4})=(m+n)[3]_{6}=m[3]_{6}+n[3]_{6}=f([m]_{4})+f([n]_{4}).$

By (1) we see that $\ker f=\{[0]_{4},[2]_{4}\}<\mathbb{Z}_{4}$ and $\operatorname{im}f=\{[0]_{6},[3]_{6}\}<\mathbb{Z}_{6}$.

Answer:

1. the identity homomorphism, $\ker f=\mathbb{Z}_{4}$ and $\operatorname{im}f=\{[0]_{6}\}$;

2. $f([1]_{4})=[3]_{6}$, $\ker f=\{[0]_{4},[2]_{4}\}$ and $\operatorname{im}f=\{[0]_{6},[3]_{6}\}$.

$\Box$