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:
- 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?
- 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.
Best Answer
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.