No simple group of order $160$, and the structure of groups of that order

Let \(|G|=160=2^5\cdot5\), and let \(P\) be a Sylow \(2\)-subgroup. Then \[ [G:P]=5. \] The action of \(G\) on the five left cosets of \(P\) gives \[ \rho:G\to S_5. \] If \(G\) were simple, the kernel of this nontrivial action would be trivial, so \(G\) would embed in \(S_5\). But \[ |G|=160>120=|S_5|, \] which is impossible. Hence no group of order \(160\) is simple.

More generally, every group of order \(160\) is solvable by Burnside’s \(p^aq^b\) theorem. Sylow gives \[ n_5\in\{1,16\}. \] If \(n_5=1\), the unique Sylow \(5\)-subgroup \(C_5\) is normal. For any Sylow \(2\)-subgroup \(P\), we then have \[ G=C_5P, \qquad C_5\cap P=1, \] so \[ G\cong C_5\rtimes P, \] with \(|P|=32\) and the action factoring through \[ \operatorname{Aut}(C_5)\cong C_4. \] Thus the strongest uniform structural conclusion is solvability; in the normal-Sylow-\(5\) case one obtains the displayed semidirect product.