Element orders in a finite quotient occur in the original group ($\varphi :G\twoheadrightarrow H,o(h)=m⟹\exists x\in G:o(x)=m$)

Let $G$ be a finite group and $\varphi :G\to H$ a surjective group homomorphism. Choose a preimage $g$ of $h$. Since $\varphi (g{)}^{o(g)}=e$, the order of a group element $m=o(h)$ divides $o(g)$; write $o(g)=mr$. Then $g^r$ has order $m$, because $(g^r{)}^s=e$ holds exactly when $mr\mid rs$, or equivalently $m\mid s$. This proves existence of an element of order $m$ in $G$. It need not be a preimage of $h$: for the quotient map ${\mathbb{Z}}_4\to {\mathbb{Z}}_2$, both preimages of the nonidentity target element have order $4$, while the order-$2$ element of the source maps to zero. Thus the assertion concerns element-order existence, not splitting of the quotient map.