between the group $(GL_2(\mathbb{F_p}),\times)$ and a group of the form ($(\mathbb{Z}/k\mathbb{Z})^{*},\times)$.
I already know that $\mid GL_2(\mathbb{F_p}) \mid=(p^2-p)(p^2-1)$.
Moreover I remark that $\mid(\mathbb{Z}/ p^2\mathbb{Z})^ *\mid=p^2-p$.
So maybe we can find a product of groups whose cardinality is the same. I don't know.
Thanks in advance !
It would never work, because $GL_2(\mathbb{F}_p)$ is non-abelian, and the other group, the group of units of the ring $\mathbb{Z}/k$, is abelian. So they cannot be isomorphic for any $p$ and any $k$.
As for the cardinalities; yes, $\phi(k)$ can indeed coincide with $(p^2-p)(p^2-1)$. For $p=2$ we have $GL_2(\mathbb{F}_2)\cong S_3$ with $6$ elements, and of course $\phi(7)=6$ too. Thus $(\mathbb{Z}/7)^{\ast}\cong C_6$ has also $6$ elements. And as you know, $S_3$ and $C_6$ are not isomorphic. Also, for $p=3$, $|GL_2(\mathbb{F}_3)|=48=|(\mathbb{Z}/105)^*|$.