Order of zero in $\mathbb{Z}_m$

Question

Would it be correct to say that the order of zero in $\mathbb{Z}_m$ under addition is infinite?

And if it's infinite then there can be no bijection between a dihedral group and $\mathbb{Z}_m$?

Answer 1

The order of an element $g$ is the least positive integer $k$ such that $g^k=e$ (using multiplicative notation). In your case $\underbrace{0}_{1~\text{time}} = 0$ so it is order $1$

Compare this to, say, the element $4$ in $\mathbb{Z}_{12}$ where $\underbrace{4+4+4}_{3~\text{times}}=0$ so the order of $4$ in $\mathbb{Z}_{12}=3$

Answer 2

The order of an element $g$ is the smallest positive integer $k$ such that $kg=0$. In this case the smallest such $k$ is $1$.

Another way to define order is as the order of the subgroup generated by $g$, denoted $\langle g \rangle $. In this case $\langle 0 \rangle=\{0\}$ which has order $1$.