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Dec 22, 2015 at 10:06 comment added YCor An improvement is to precisely give $D(G)$ for any abelian (or nilpotent) $G$. Namely, let $G$ be nilpotent and let $G'$ be its quotient by its Frattini subgroup, so $G'$ is a product of $k$ cyclic groups of prime order. Then $D(G)=D(G')=k$.
Dec 22, 2015 at 9:33 history edited Geoff Robinson CC BY-SA 3.0
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Dec 22, 2015 at 0:47 history edited Geoff Robinson CC BY-SA 3.0
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Dec 21, 2015 at 23:13 history edited Geoff Robinson CC BY-SA 3.0
Expanded answer on when maximal value is attained.
Dec 21, 2015 at 15:53 comment added Geoff Robinson I read the question differently, but the PO may clarify. I think it is a rather tall order to expect to know $D(G)$ precisely for all finite groups, but I am prepared to be proved wrong.
Dec 21, 2015 at 15:47 comment added Seva What if $G$ is not abelian of square-free exponent? As I understand if, the PO wants to know $D(G)$ precisely for all finite groups $G$.
Dec 21, 2015 at 15:39 history edited Geoff Robinson CC BY-SA 3.0
typos
Dec 21, 2015 at 15:26 history answered Geoff Robinson CC BY-SA 3.0