Give an example of a non abelian group of order $55$.
To find non abelian group the simplest way is to find one non abelian group whose order divides the order of given group and then we take the group which is the external direct product of the non abelian group and some other abelian group.
For example to find a non abelian group of order $36$ we take the permutation group $S_3$ and take the group $S_3\otimes \Bbb Z_6$. But using this way we can not have a group of order $55$ since any group of order $5$ or $11$ will be abelian.
So how do we proceed?
Best Answer
Notince that $\operatorname{aut}\Bbb Z_{11}\cong \Bbb Z_{11}^*\cong \Bbb Z_{10}$. Therefore you can consider an element $g\in\operatorname{aut}\Bbb Z_{11}$ of order $5$ and the homomorphism $\phi:\Bbb Z_5\to \operatorname{aut}\Bbb Z_{11}$, $\phi(m)=g^m$. Then you have a non-abelian semidirect product $\Bbb Z_{11}\rtimes_\phi \Bbb Z_5$ with operation $(a,b)*(c,d)=(a+\phi(b)(c),b+d)$.