I know this is an old quetion, but I've certainly been disappointed with the given answers. The question is: There exists a characterization of the natural numbers $n$ such that there exist at least one non-abelian group of order $n$? In this web http://oeis.org/A060652, those numbers are called "non-abelian orders" and they state this:

Let the prime factorization of $n$ be $p_1^{e_1}\cdots p_r^{e_r}$. Then $n$ is in this sequence (of non-abelian orders) if $e_i>2$ for some $i$ or $p_i^k\equiv1$(mod $p_j$) for some $i$ and $j$ and $1\leq k\leq e_i$.

I've been trying to figure out a proof and this is what I've got:

Let $n=p_1^{e_1}\cdots p_r^{e_r}$, then there exist an non-abelian group of order $n$ if $e_i>2$ for some $i$ or $p_i\equiv 1$(mod $p_j$) for some $i\neq j$.

It is clear that if both of these conditions fail to hold, then the only remainig option is $p_i^2\equiv 1$(mod $p_j$) for some $i\neq j$. Then the problem "reduces" to this:

  1. If $p_i^2\equiv 1$(mod $p_j$) for some $i\neq j$, then there exist at least one non-abelian group of order n?
  2. If 1 is true, does the converse of the theorem also holds? i.e. If there exists a non-abelian group of order $n=p_1^{e_1}\cdots p_r^{e_r}$, then $e_i>2$ for some $i$ or $p_i^k\equiv1$(mod $p_j$) for some $i$ and $j$ and $1\leq k\leq e_i$.

Thanks in advance.

  • $\begingroup$ Presumably when you say "there exist[s] an abelian group" you mean "there exists a nonabelian group." $\endgroup$ – Qiaochu Yuan Apr 4 '15 at 19:21
  • $\begingroup$ Yes, that's It. Thanks. $\endgroup$ – Jose Paternina Apr 4 '15 at 19:23

For every prime $p$ there is a nonabelian group of order $p^3$, namely the Heisenberg group $H(\mathbb{F}_p)$, or equivalently the semidirect product

$$\mathbb{Z}_p^2 \rtimes \mathbb{Z}_p$$

where $\mathbb{Z}_p$ is the cyclic group of order $p$ and the action is by multiplication by $\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]$. By taking direct products of such groups with cyclic groups we can construct a nonabelian group of order $n = \prod p_i^{e_i}$ whenever some $e_i \ge 3$.

Now suppose that each $e_i \le 2$. Then every group of order $p_i^{e_i}$ is abelian, and hence every Sylow subgroup of a group of order $n$ is abelian, so we need to do a construction involving at least two primes. If $p_i^k \equiv 1 \bmod p_j$ for some $1 \le k \le e_i$, then the cyclic group $\mathbb{Z}_{p^j}$ acts on $\mathbb{Z}_{p_i}^k$ as follows: we can identify $\mathbb{Z}_{p_i}^k$ with the underlying abelian group of $\mathbb{F}_{p_i^k}$, whose unit group has order $p_i^k - 1$, which by hypothesis is divisible by $p_j$. Hence there is a unit of order $p_j$, and $\mathbb{Z}_{p^j}$ acts by multiplication by this unit. Now we can take the semidirect product

$$\mathbb{Z}_{p_i}^k \rtimes \mathbb{Z}_{p_j}$$

which is a nonabelian group of order $p_i^k p_j$, and since this divides $n$, we can again take direct products with cyclic groups and construct a nonabelian group of order $n$.

As for the converse see this math.SE answer.

  • $\begingroup$ Thanks for the reply, It was a nice approach. $\endgroup$ – Jose Paternina Apr 4 '15 at 19:36

Dickson proved that only abelian groups exist of order n iff $n=p_1...p_nq_1^2...q_m^2$ where $n$ is relatively prime to $n=(p_1-1)...(p_n-1)(q_1^2-1)...(q_m^2-1)$ so the complement of this in the set of natural numbers gives you the answer.


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