Prove that $m$ divides the characteristic of $R.$

Question

I want to answer this question:

Let $R$ be a ring (with $1$), and let $H$ be the cyclic subgroup of $(R,+)$ generated by $1.$ The characteristic of $R$ is $n$ if $H\cong \mathbb{Z}_{n},$ and is $0$ if $H \cong \mathbb{Z}.$\ (a) Show that, if $a \in R$ has additive order $m,$ then $m$ divides the characteristic of $R.$

My question is:

I get this hint to answer it: If $H \cong \mathbb{Z}$, this is obvious (why?). So assume $H\cong \mathbb{Z}_{n}$ for some $n$. Remind yourself that $(R,+)$ is an abelian group. Use the following fact from elementary group theory.

Lemma: Suppose that $G $is a group with identity $e$, and $x$ in $G$ has finite order t. If $x^s=e$ for some integer $s$, then $t$ divides $s$.

But I do not understand how this hint will lead me to prove that $m$ divides the characteristic of $R.$ could anyone help me in answering this question, please?

Answer 1

Case 1: The characteristic of $R$ is $0$.
Clearly, $m$ divides $0$ so this is obvious.

Case 2: The characteristic of $R$ is $n$ where $n\neq 0$.
Note that $$na=n(1a)=(n1)a=0a=0.$$ Since $m$ is the additive order of $a$, by the Lemma you stated, $m$ must divides $n$.