 Solution

Let $n=o(g)$. A group homomorphism preserves powers and the identity element, soThe preceding divisibility property givesNow assume $\varphi$ is surjective and choose $h\in H$ of order $m$. Take a preimage $g$ of $h$, and write $o(g)=mr$, as the divisibility result permits. The element $g^r$ has order exactly $m$: its $m$th power is $e$, while $(g^r{)}^s=e$ implies $mr\mid rs$, hence $m\mid s$. Therefore $G$ contains an element of order $m$, namely $g^r$.

The preimage $g$ itself need not have order $m$; taking the suitable power is the essential step. The constructed element need not map to the original $h$. This proves that element orders in a finite quotient occur in the original group.