# Are there groups with conjugacy classes as large as the divisors of perfect numbers?

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

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