How do you prove a group of order 27 can be non abelian?
From what I tried:
Say $|G|=27$ then it's a p-group therefore the ceneter is non trivial so $|Z(G)|$ could be either $3,9,27$ if $|Z(G)|=27$ then the group is abelian. if $|Z(G)|=9$ then from lagrange $|G/Z(G)|=3$ therefore cyclic therefore $G$ is abelian (which is very strange to me although I'm using the theorems I just found an abelian group $G\neq Z(G)$). and if $Z(G)=3$ then I know that $G/Z(G)$ is abelian because it's of order 9 but I don't know how to continue from here.
Any help will be appreciated.
Best Answer
You should prove by giving an example since not every group of order $27$ are non-abelian.
Take $$G=\{\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\;|\; a,b,c\in \Bbb{Z}_3\}$$ Try to verify that this group is a non-abelian group of order $27$.
In fact it is called the Heisenberg group over $\Bbb{Z}_3$.
It is subgroup of $GL(n,\Bbb{Z}_3$)