Proof: $\operatorname{exp}G=o(g)$ for a finite abelian group ($g$ maximal order)

Question

I don't understand why $h^{o(g)}\neq 1$ implies $f_i>e_i$, and not just $f_i\neq e_i$. It seems rather counterintuitive to me that $f_i>e_i$, given that $o(h)\leq o(g)$. Could someone clarify this step for me?

Answer 1

Suppose it's not true: then, for every $i$, $f_i\le e_i$, so $o(h)$ is a divisor of $o(g)$, and $h^{o(g)}=1$, contrary to the hypothesis.