For a finite group G, let |G| denote the order of G and write $D(G) = \sum_{N \triangleleft G} |N|$, the sum of the orders of the normal subgroups. I would like to call G "perfect" if D(G) = 2|G|, since then the cyclic group of order n is perfect if and only if the number n is perfect. But the term "perfect group" is taken, so I'll call such a group **immaculate**.

My question is:

Does there exist an immaculate group of odd order?

Since the *cyclic* immaculate groups correspond one-to-one with the perfect numbers, a "no" answer would immediately prove the famous conjecture that there are no odd perfect numbers. However, perhaps someone can easily see that there is a *non*-cyclic immaculate group of odd order, proving that the answer is "yes".

Here's what I know. There are no abelian immaculate groups except for the cyclic ones. (*Edit*: more generally, if $D(G) \leq 2|G|$ then every abelian quotient of $G$ is cyclic. Proof: not hard, and given here.) However, there do exist nonabelian immaculate groups, e.g. $S_3 \times C_5$ (of order 30). Derek Holt has computed all the immaculate groups of order less than or equal to 500. Their orders are
$$
6, 12, 28, 30, 56, 360, 364, 380, 496
$$
(Integer Sequence A086792). Of these, only 6, 28 and 496 are perfect numbers; the rest correspond to nonabelian immaculate groups. Some nonabelian immaculate groups of larger order are also known, e.g. $A_5 \times C_{15128}$, $A_6 \times C_{366776}$, and, for each even perfect number n, a certain group of order 2n. But these, too, all have even order.

*Edit*: Steve D points out that p-groups can never be immaculate. This also appears as Example 2.3 here; it follows immediately from Lagrange's Theorem. I should have mentioned this, as it rules out an easy route to a "yes" answer.

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