Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T06:27:33Z http://mathoverflow.net/feeds/question/54851 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54851/is-there-an-odd-order-group-whose-order-is-the-sum-of-the-orders-of-the-proper-no Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups? Tom Leinster 2011-02-09T05:50:03Z 2011-02-21T09:56:06Z <p>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 <a href="http://en.wikipedia.org/wiki/Perfect_group" rel="nofollow">taken</a>, so I'll call such a group <strong>immaculate</strong>.</p> <p>My question is:</p> <blockquote> <p>Does there exist an immaculate group of odd order?</p> </blockquote> <p>Since the <em>cyclic</em> 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 <em>non</em>-cyclic immaculate group of odd order, proving that the answer is "yes".</p> <p>Here's what I know. There are no abelian immaculate groups except for the cyclic ones. (<em>Edit</em>: more generally, if $D(G) \leq 2|G|$ then every abelian quotient of $G$ is cyclic. Proof: not hard, and given <a href="http://arxiv.org/abs/math.GR/0104012" rel="nofollow">here</a>.) 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 $$6, 12, 28, 30, 56, 360, 364, 380, 496$$ (<a href="http://oeis.org/A086792" rel="nofollow">Integer Sequence A086792</a>). 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.</p> <p><em>Edit</em>: Steve D points out that p-groups can never be immaculate. This also appears as Example 2.3 <a href="http://arxiv.org/abs/math.GR/0104012" rel="nofollow">here</a>; it follows immediately from Lagrange's Theorem. I should have mentioned this, as it rules out an easy route to a "yes" answer.</p> http://mathoverflow.net/questions/54851/is-there-an-odd-order-group-whose-order-is-the-sum-of-the-orders-of-the-proper-no/54910#54910 Answer by Tom De Medts for Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups? Tom De Medts 2011-02-09T17:57:03Z 2011-02-09T17:57:03Z <p>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...</p> http://mathoverflow.net/questions/54851/is-there-an-odd-order-group-whose-order-is-the-sum-of-the-orders-of-the-proper-no/55026#55026 Answer by François Brunault for Is there an odd-order group whose order is the sum of the orders of the proper normal subgroups? François Brunault 2011-02-10T12:58:41Z 2011-02-21T09:56:06Z <p>I did a little computer search and I think I found an example of an odd immaculate group.</p> <p>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</p> <p>\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 :</p> <p>\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 $|G| = 355433039577 = 3^4 \cdot 7 \cdot 11^2 \cdot 19^2 \cdot 113 \cdot 127$.</p> <p>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 $4$.</p> <p>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).</p> <p>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 $m$.</p>