= Solution
A <normal subgroup> $H$ satisfies $gHg^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. A left <coset> is $gH=\{gh:h\in H\}$ and a right coset is $Hg=\{hg:h\in H\}$.
If $|G|=2|H|$, then for $g\in H$ both <cosets> equal $H$. For $g\notin H$, neither coset intersects $H$: an intersection would imply $g\in H$. Both have $|H|$ elements, because multiplication by $g$ is a <bijection>. They therefore both equal $G\setminus H$, proving $gH=Hg$ in this case too. This establishes that an <index-two subgroup is normal>.
For <Lagrange's theorem>, left <cosets> partition $G$. Indeed if $g_1h_1=g_2h_2$, then $g_1H=g_2H$ by multiplying by elements of $H$; otherwise the <cosets> are disjoint. Every coset has $|H|$ elements, so if there are $[G:H]$ <cosets>,
$$
\boxed{|G|=[G:H]|H|.}
$$
The <group order> of $H$ thus divides the <group order> of $G$. Normality is not needed for this divisibility statement.
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