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](http://en.wikipedia.org/wiki/Perfect_group), 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](http://arxiv.org/abs/math.GR/0104012).) 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](http://oeis.org/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](http://arxiv.org/abs/math.GR/0104012); it follows immediately from Lagrange's Theorem. I should have mentioned this, as it rules out an easy route to a "yes" answer.



Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups?
I did a little computer search and I think I found an example of an odd immaculate group.


I searched for groups of the form $G=(C\_q \rtimes C\_p) \times C\_N$ with odd primes $p,q$ such that $p | q-1$ and $N$ an odd integer satisfying $(N,pq)=1$. Using Tom's notations and results, we have


\begin{equation\*}
\frac{D(G)}{|G|} = \frac{D(C\_q \rtimes C\_p)}{|C\_q \rtimes C\_p|} \cdot \frac{D(C\_N)}{|C\_N|} = \frac{1+q+pq}{pq} \cdot \frac{\sigma(N)}{N}
\end{equation\*}
where $\sigma(N)$ denotes the sum of divisors of $N$. We want $\frac{\sigma(N)}{N} = \frac{2pq}{1+q+pq}$. Since the last fraction is irreducible, $N$ has to be of the form $N=(1+q+pq)m$ with $m$ odd. I found the following solution :


\begin{equation\*}
p=7, \quad q=127, \quad m=393129.
\end{equation\*}
This gives the immaculate group $G=(C\_{127} \rtimes C\_7) \times C\_{399812193}$, which has order $|G| = 355433039577 = 3^4 \cdot 7 \cdot 11^2 \cdot 19^2 \cdot 113 \cdot 127$.


Edit : here is the beginning of an explanation of "why" $m$ is square in this example ($393129=627^2$). Recall that an integer $n \geq 1$ is a square if and only if its number of divisors is odd (consider the involution $d \mapsto \frac{n}{d}$ on the set of divisors of $n$). If $n$ is odd, then all its divisors are odd, so that $n$ is a square if and only if $\sigma(n)$ is odd. Now consider $N=(1+q+pq)m$ as above. The condition on $\sigma(N)/N$ implies that $\sigma(N)$ is even but not divisible by $4$.


If we assume that $1+q+pq$ and $m$ are coprime, then $\sigma(N)=\sigma(1+q+pq) \sigma(m)$, so the reasoning above shows that $1+q+pq$ or $m$ is a square (but not both). If $1+q+pq=\alpha^2$ then $\alpha \equiv \pm 1 \pmod{q}$ so that $\alpha \geq q-1$, which leads to a contradiction. Thus $m$ is a square (it is possible to show further that $1+q+pq$ is a prime times a square).


If $1+q+pq$ and $m$ are not coprime, the situation is more intricate (this is what happens in the example I found : we had $\operatorname{gcd}(1+q+pq,m)=9$). Let $m'$ be the largest divisor of $m$ which is relatively prime to $1+q+pq$. Put $m=\lambda m'$. Then $\lambda(1+q+pq)$ or $m$ is a square. I don't see an argument for excluding the first possibility, but at least if $\lambda$ is a square then so is $m$.


