Prove that if G is a finite group, and N is a normal subgroup of G,
then for any x Є G, O(x) in G is divisible by order of Nx in G/N.
If $o(x) = k$ we surely have that: $x^k = e$, therefore:
$(Nx)^k = N(x^k) = Ne = N$, the identity of $G/N$. Hence $o(Nx)$ must divide $k$, for if not, say: $o(Nx) = m$ with:
$k = qm + r$ and $0 < r < m$, we have:
$N = (Nx)^k = (Nx)^{qm+r} = (Nx)^{qm}(Nx)^r = ((Nx)^m)^q(Nx)^r = N^q(Nx)^r = N(Nx)^r = (Nx)^r$
contradicting the fact that $m$ is the least positive integer with $(Nx)^m = N$.