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First of all the $k(G)$ cannot be smaller than the size of any proper subgroup of $G$, because if $H$ is a proper subgroup, $gH$ is a coset, $g\not\in H$, then $gHgH$ does not contain intersect $gH$ (it will not contain if $g$). ghgh'=gh''$, then$g\in H$). For example, if$G$is Abelian,$|G|$is not prime (i.e.$G$is not cyclic of prime order), then$k(G)>\sqrt{|G|}$: look at the size of a maximal proper subgroup. 1 First of all the$k(G)$cannot be smaller than the size of any proper subgroup of$G$, because if$H$is a proper subgroup,$gH$is a coset,$g\not\in H$, then$gHgH$does not contain$gH$(it will not contain$g$). For example, if$G$is Abelian,$|G|$is not prime (i.e.$G$is not cyclic of prime order), then$k(G)>\sqrt{|G|}\$: look at the size of a maximal proper subgroup.