Is a group of order 1280 simple or not?

Question

Show whether or not $G$ is simple If $|G| = 1280$ and explain your reasoning.

I am told I have to use Sylow's theorems.

I have begun by finding that $n_{2}\in \{1,5\}$ and $n_5\in \{1,256\}$. I then know that if $n_2=1$ or $n_5=1$, then $G$ would not be simple. So I am trying to see if $n_2=5$ AND $n_5=256$ but am struggling with how to do this part.

Can I say that if $G$ is simple, $|G| =1280$ must divide 5!, which is not the case?

Answer 1

Sylow's theorems say that Sylow $p$-subgroups are conjugate to each other. This gives a group action of $G$ on the set of Sylow $p$-subgroups which is transitive. In other words you get a group morphism from $G$ to $S_{n_p}$ which is not the trivial map as the action is transitive. What are the possibilities for the kernel of this map if $G$ were simple?

Answer 2

The cyclic group $\mathbf Z/1280\mathbf Z$ is a counter-example, as as an abelian group is simple if and only if it has prime order.