= Solution
{c}
Let $n=o(g)$. A <group homomorphism> preserves powers and the <identity element>, so
$$
\phi(g)^n=\phi(g^n)=\phi(e)=e_H.
$$
The preceding divisibility property gives
$$
\boxed{o(\phi(g))\mid o(g).}
$$
Now assume $\phi$ 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 \b[$G$ 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>.
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