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Are there groups, where the conjugacy classes have elements according to the divisors of perfect numbers larger than $6$?

I already got the first example, so therefore $6$ is excluded: $S_3$.

The corresponding GroupProps page on groups of order $28$ is still empty, but OEIS/A000001 says that there are $4$ groups having $28$ elements.

Any idea?

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Can you explain your question a little bit better? What does it mean "elements according to the the divisors of perfect numbers"?? – user641 Sep 16 '13 at 21:49
@SteveD let me work out the "group of order 28" example: I'm asking if there is group with conjugacy classes having $1,2,4,7$ and $14$ elements. – draks ... Sep 16 '13 at 21:51
Oh ok: then the answer is no. If $4$ divides the order of the group $G$ (let's say $|G|=n$), then there cannot be an element whose conjugacy class has size $n/2$. This is because such an element would be order $2$ and self-centralizing, which is impossible (normalizers grow in p-groups). – user641 Sep 16 '13 at 21:53
@SteveD so 6 is the only example...? – draks ... Sep 16 '13 at 21:56
@draks... As far as I know it's not proven that there are no odd perfect numbers. – Daniel Fischer Sep 16 '13 at 21:58
up vote 1 down vote accepted

Suppose $|G|=n$, and that $n$ is a perfect number, with $G$ having a conjugacy class of every possible size. Let's show $n=6$.

First, let $p$ be the smallest prime dividing $n$, and let $x\in G$ be an element with conjugacy class size $n/p$. Then $C_G(x)$ has size $p$, so $x$ is an element of order $p$ that is self-centralizing. Since $N_G(\langle x\rangle)/C_G(\langle x\rangle)$ has order dividing $p-1$, the minimality of $p$ implies $N_G(\langle x\rangle)=C_G(\langle x\rangle)$. Thus $G$ has a normal $p$-complement, so that $G=H\rtimes \langle x\rangle$, with $|H|=n/p$.

Now let $y\in H$ have a conjugacy class size of $p$. Then the centralizer of $y$ has order $|H|$, so $y\in Z(H)$. This means that if $|y|=k$, then every element in $H$ has centralizer at least of order $k$, so conjugacy class size dividing $n/k$. Since we want conjugacy classes of every possible size, this is only possible if $n/k=p$, or $n=kp$. In particular, $H$ is cyclic with generator $y$.

But then the only possible conjugacy classes are those in $H$ of size $p$, and those outside $H$ of size $n/p=k$. Thus $k$ must be a prime $q$, and $n=pq$. By the assumption that $n$ is perfect, $pq=p+q+1$. Since $p<q$, working mod $q$ shows $$ p+1=q.$$

Thus $p=2$, $q=3$, and $n=6$.

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This argument also shows $G$ must be $S_3=C_3\rtimes C_2$. – user641 Oct 5 '13 at 8:31
thanks, it just took me some time to get through the first part... – draks ... Oct 10 '13 at 20:04

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