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 $m$.

Almost a decade late to the party, but here is another example:

$$(C_7 \rtimes C_{3^2}) \times (C_{19^2} \rtimes C_5) \times C_{11^2 \cdot 197} = \text{SmallGroup}(63, 1) \times \text{SmallGroup}(1805, 2) \times C_{23837}.$$ where $\text{SmallGroup}(a,b)$ denotes the $b$th group of order $a$ according to the database of GAP. Then: $$\frac{D(C_7 \rtimes C_{3^2})}{|C_7 \rtimes C_{3^2}|}\frac{D(C_{19^2}\rtimes C_5)}{|C_{19^2}\rtimes C_5|}\frac{D(C_{11^2})}{|C_{11^2}|}\frac{D(C_{197})}{|C_{197}|} = \frac{95}{63}\cdot \frac{2167}{1805}\cdot \frac{133}{121}\cdot \frac{198}{197} $$ $$=\frac{5 \cdot 19}{3^2\cdot 7} \cdot\frac{11 \cdot 197}{5\cdot19^2}\cdot \frac{7 \cdot 19}{11^2}\cdot \frac{2 \cdot3^2\cdot 11}{197}=2.$$ This group is of order $2710624455$ and is thus about 100 times smaller than François Brunault's example.

I found this group for a project for one of my courses. Immaculate groups seem to be very rare in general, examples of order are even harder to find. For further results, see http://www.math.ru.nl/~bosma/Students/JorisNieuwveld/A_note_on_Leinster_groups.pdf

Immaculate groups of odd order do exist; for example (C13 : C3) x C477, a group of order 18603. In fact, I happen to be writing a paper with Attila Maróti precisely about immaculate groups...