Good evening, first of all sorry for my bad english and for my bad knowledge of algebra; so I need help to solve this exercise please.
I want to show my idea (probably wrong):
like in another Answer in this forum, I know that a group $G^*$ of order $765$ is abelian; so I observe that $|G|=|G^*|\cdot 3$ and the quotient $G/C_3$, that has order $765$, is an abelian group; I think that $C_3$ is the minimal subgroup of $G$ with this property, then I may assert that $G'=C_3$?
If is correct, I suspect that $G$ is metabelian but for say this, I need to show that the down central series defined by:
$G=\Gamma_0 \supseteq \Gamma_1\supseteq\ldots\ldots\supseteq\Gamma_n=e$
with $\Gamma_i=[G,\Gamma_{i-1}]$
has lenght $2$; so is like:
$G\supseteq G'\supseteq e$
or
$G\supseteq C_3 \supseteq e$
so I'm not able to show that $[G,C_3]=e$ and hence $C_3=G'=Z(G)$...
Please UNDERLINE mercilessly all my mistakes.
With regard,
Leonardo Ventura.