How would you start classifying groups of order \(56\)? Is there a nontrivial action of \(C_7\) on a group of order \(8\)?
Solution
Let \(|G|=56=8\cdot7\). Sylow gives \[ n_7\equiv1\pmod7,\qquad n_7\mid8, \] so \(n_7=1\) or \(8\).
If \(n_7=1\), the Sylow \(7\)-subgroup is normal. If \(n_7=8\), the eight Sylow \(7\)-subgroups contribute \[ 8(7-1)=48 \] nonidentity elements. Exactly eight elements remain, and every Sylow \(2\)-subgroup has eight elements, so there is a unique Sylow \(2\)-subgroup. Thus in every group of order \(56\), at least one Sylow subgroup is normal, giving the starting semidirect-product reduction.
Now let \(P\) be a group of order \(8\). A nontrivial action of \(C_7\) on \(P\) exists exactly when \(\operatorname{Aut}(P)\) contains an element of order \(7\). For the five groups of order \(8\): \[ \begin{array}{c|c} P&|\operatorname{Aut}(P)|\\ \hline C_8&4\\ C_4\times C_2&8\\ D_4&8\\ Q_8&24\\ C_2^3&|\operatorname{GL}_3(\mathbb F_2)|=168 \end{array} \] Only \(C_2^3\) admits such an action. Since Sylow \(7\)-subgroups of \(\operatorname{GL}_3(\mathbb F_2)\) are conjugate, all faithful actions are equivalent up to automorphisms of \(C_2^3\) and \(C_7\). Hence there is a unique nontrivial semidirect product \[ C_2^3\rtimes C_7 \] of this form.