# Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups?

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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It is not hard to see for primes p, non-cyclic p-groups cannot be immaculate. Don't know if that helps or not (I was trying to construct an odd-order immaculate group using that odd order groups are solvable, but there extensions make things even harder). – Steve D Feb 9 '11 at 7:17
It took me only a few minutes to run through the small groups library and confirm that there are no odd order immaculate groups of order up to 2000. Following the comments above, I did not check groups of prime power order. – Derek Holt Feb 9 '11 at 9:18
A remark about the terminology. I once read that "perfect number" might not be the correct translation of the original word "Τέλειος ἀριθμός" which can be found in Euclid's Elements. According to the French wiktionary, one translation of the ancient greek "Τέλειος" is indeed "immaculate", other possibilities being "accomplished" or "complete" (please correct my English). Would an Hellenist confirm these facts ? – François Brunault Feb 9 '11 at 11:57
Those are all very good translations... as confirmed by the scholar of ancient Greek currently next to me. The idea of being "complete" was related to being "perfect" in the time of Plato. Later languages like Coptic use the same terminology. XHK EBOL means both complete and perfect in Coptic, and the noun form means "end", just like "TELEIOS". – Marty Feb 9 '11 at 16:45
Are there any perfect immaculate groups? There are none in the GAP perfect groups library. – Jack Schmidt Feb 10 '11 at 0:13

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 $m$.

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Is there any interesting reason why $m$ is a perfect square (here $m=627^2$) ? – François Brunault Feb 10 '11 at 13:18
Bravo ! – Tom Leinster Feb 10 '11 at 15:48
And is there any interesting reason why p and q are Mersenne primes? – Tom Leinster Feb 10 '11 at 15:51
@Gerhard : thanks for mentioning this result on odd perfect numbers which I didn't know. It would be indeed interesting to prove something similar here. Note that in my setting $N=m(1+q+pq)$ has to be deficient (nevertheless, the ratio $\sigma(N)/N$ should be close to 2, in my example it is $1.748\ldots$). – François Brunault Feb 10 '11 at 17:21
So the example by Tom De Medts almost fits in with this answer, except that gcd(N,pq) there is 3, not 1... – Max Horn Feb 10 '11 at 17:33

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...

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Sorry, but I think this is wrong. Your group may be written as $(C_{13} : C_3) \times C_9 \times C_{53}$. There are 3 different homomorphisms $\phi\colon (C_{13}:C_3) \to C_9$. For each such $\phi$ you get a different normal subgroup $\{(a,b,c)\in (C_{13}:C_3)\times C_9 \times C_{53} \mid a\phi = b^3\}$ of order $39 \cdot 3 \cdot 53$, in total 3 normal subgroups of order $18603/ 3$. And there are others, of course. Alternatively, use GAP: g:=DirectProduct( SmallGroup( 39,1 ), SmallGroup( 477, 1 ) );; Sum( NormalSubgroups( g ), Size ); 54054 – Frieder Ladisch Feb 9 '11 at 19:28
What does this $G:H$ mean? – Martin Brandenburg Feb 9 '11 at 20:14
@Martin: Here it means $C_3$ acting nontrivial on $C_{13}$. This defines the group up to isomorphism. I don't know who introduced this notation, but it is used in the ATLAS of finite groups. If I remember right, they use $A_6:2_1$, $A_6:2_2$ and $A_6:2_3$ to denote the three nonisomorphic groups containing $A_6$ nontrivially as normal subgroup of index 2. – Frieder Ladisch Feb 9 '11 at 20:55
Maybe a better way to see that this group can not be immaculate is this: if $G$ has a normal subgroup $N$ such that $G/N\cong C_p\times C_p$ then $G$ has $p+1$ normal subgroups of order $|G|/p$, so it is abundant. In particular, for a non-abundant group, $G/G'$ must by cyclic, which is not the case here. – Frieder Ladisch Feb 9 '11 at 21:17
Sorry for having posted this incorrect example! (It was due to a silly mistake in my computer program where I was mixing up normal subgroups in a composition series with their factors, and I was too careless not to double-check the example.) – Tom De Medts Feb 10 '11 at 21:19