= Element orders in a finite quotient occur in the original group
{title2=$\phi:G\twoheadrightarrow H,\ o(h)=m\ \Longrightarrow\ \exists x\in G:\ o(x)=m$}
Let $G$ be a <finite group> and $\phi:G\to H$ a surjective <group homomorphism>. Choose a preimage $g$ of $h$. Since $\phi(g)^{o(g)}=e$, 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.
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