$\{0\}\times 3\mathbb{Z}/9\mathbb{Z}$ as a characteristic subgroup of $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$

Question

Reading a solution of an exercise, I found the statement

We can say that $\{0\} \times 3\mathbb{Z}/9\mathbb{Z}$ is a characteristic subgroup of $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$ because is the kernel of the multiplication by $3$.

I suppose that "kernel of the multiplication by $3$" means that if $\varphi :\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z} \longrightarrow \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z}$ and $\varphi (a,b)=(3a,3b)=(0,3b)$, then $\{0\} \times 3\mathbb{Z}/9\mathbb{Z}=\ker \varphi$, but I don't understand how this implies that this subgroup is characteristic.

Edit (just to be clear): $3\mathbb{Z}/9\mathbb{Z}=\langle 3 \rangle =\{0,3,6\}$

Answer 1

If $G$ is an abelian group the kernel of the "multiplication" by $n$ morphism is characteristic. This follows from the fact that if $f$ is an authomorphism of $G$ then $f(ng)=nf(g)$ for $g\in G$, so if $ng=0$ then $nf(g)=0$ (the other direction follows applying $f^{-1}$). In other words $f$ sends "$n$th roots of unity" to "$n$th roots of unity".