If $|a|= \infty$, then $$ is infinite subgroup I'm trying to understand the proof of:


  Let $G$ be group and $a \in G$.
  If $|a|= \infty$, then $&lt;a&gt;$ is infinite subgroup.


The proof I have here goes like:


  Let $|a|=\infty$. Then $a^i ≠ a^j$ for all $i≠j$, because otherwise
  $|a|≤ |i-j|.$ Which proves the statement.




I don't understand why must $a^i ≠ a^j$ and how this relates to the order of the element (or the subgroup).
Suppose that for some $i\not= j$, you have $a^{i}=a^{j}$. Assume without loss of generality that $i&lt;j$. Then, multiplying $a^{-i}$ to both sides gives $e=a^{j-i}$. Since $j-i&gt;0$, this gives that $|a|\leq j-i$, contradicting that $|a|=\infty$. Therefore, it must be that for all $i\not= j$, $a^{i}\not= a^{j}$. Hence, $&lt;a&gt;=\{a^{i}\mid i\geq 0\}$ is an infinite set.
It is known that if $G$ is finite cyclic group and has order $n$ then the subgroup $H$ order will be number divisible by $n$, so let the subgroup $H$ order be $r$ then $r.x=n$ for positive $x$, now in your task assume that $H$ is finite then $n$=infinity must divide $r$ which is not true, so the subgroup $H$ is finite.