Is there a non abelian group of order 759?

Question

I tried to use Sylow theorems to prove that there is not, but it is not trivial.

Answer 1

HINT:

$$759 = 3 \cdot 11 \cdot 23$$

Show that there exists a non-commutative group of order $11 \cdot 23$.

${\bf Added:}$ Let's construct a non-commutative group $G$ of order $23\cdot 11$ as follows: the elements of $G$ are pairs $(a, m)$ where $a \in \mathbb{Z}/23$ and $m \in \mathbb{Z}/11$ and the multiplication is $$(a_1, m_1) \cdot (a_2, m_2) = ( a_1 + 2^{m_1}\cdot a_2\ ,\ m_1 + m_2)$$ where we use the fact that $2^{11} \equiv 1 \mod 23$.

Answer 2

My first instinct is to approach this problem through representation theory, and a process of elimination to show such a group cannot exist. These facts may prove useful

(1) The prime factorisation is $759 = 3 \cdot 11 \cdot 23$.

(1.5) Look at how few factors there are!

(2) Each conjugacy class is a normal subgroup, and every normal subgroup is a union of conjugacy classes. If the group is non-Abelian, some conjugacy class has more than one element. Since $759$ has only three factors, the order has only about six possibilities. Can you make a short list of what the conjugacy classes could look like?

(3) The group has as many conjugacy classes as it has irreducible representations.

Exactly what (3) means is probably not needed. The useful fact is that, if there were three irreducible representations say $\rho_1, \rho_2, \rho_3$, then each has a positive integer $d_i$ called its degree such that $d_1^2 + d_2^2 + d_3^2 = |G| = 759$.

But maybe we cannot express $759$ as a sum of exactly three cubes. This would prevent there being exactly three conjugacy classes. Perhaps a similar argument can be used for each other possibility vis-a-vie the conjugacy class layout.

Answer 3

A number $n$ has the property that every group of order $n$ is abelian (in this case we call the number an abelian number), if $n$ is cubefree and there is no prime power $p^k$ dividing $n$ and a prime $q$ dividing $n$ with $q|p^k-1$.

Here this is not the case because of $11|23-1$, so $759$ is not abelian, so there must be a non-abelian group with order $759$. There are $2$ groups of order $759$ , $1$ is abelian and $1$ is non-abelian.

Answer 4

Let the order of $G$ be 759. It is not difficult to see that the 23-Sylow subgroup is normal in $G$. Let $a$ generate the 23-Sylow subgroup in $G$. Now suppose $b\ne1$ in $G$ is such that order of $b$ is 33. If the commutation $ab=ba^2$ is assumed, then the group generated by $a$ and $b$ is of order $759$.