State Sylow’s theorems.
Classify up to isomorphism the groups of order 35.
Show that there are no simple groups of order 24.
Solution
1. The Sylow theorems have the following form.
Let \(G\) be finite, let \(p\) be prime, and write \(|G|=p^am\) with \(p\nmid m\). A Sylow \(p\)-subgroup is a subgroup of order \(p^a\). Such subgroups exist. Every \(p\)-subgroup of \(G\) is contained in a Sylow \(p\)-subgroup, and all Sylow \(p\)-subgroups are conjugate. For a Sylow \(p\)-subgroup \(P\), their number is \[ n_p=[G:N_G(P)],\qquad n_p\mid m,\qquad n_p\equiv1\pmod p, \] where \(N_G(P)=\{g\in G:gPg^{-1}=P\}\). In particular, \(P\) is normal exactly when \(n_p=1\) [@DF04].
2. Every group of order \(35\) is cyclic; hence \(C_{35}\) is theunique isomorphism class.
Since \(35=5\cdot7\), step 1 gives \[ n_7\mid5,\quad n_7\equiv1\pmod7, \qquad n_5\mid7,\quad n_5\equiv1\pmod5. \] The divisors of \(5\) are \(1,5\), and only \(1\) is \(1\) modulo \(7\). The divisors of \(7\) are \(1,7\), and only \(1\) is \(1\) modulo \(5\). Thus the subgroups \(P\) and \(Q\) of orders \(5\) and \(7\) are both normal. Their intersection is trivial, since its order divides both \(5\) and \(7\).
For \(x\in P\) and \(y\in Q\), the commutator \(xyx^{-1}y^{-1}\) lies in both normal subgroups, hence is \(1\). Thus they commute. The multiplication map \(P\times Q\to G\) is a homomorphism with trivial kernel: \(xy=1\) implies \(x=y^{-1}\in P\cap Q\). It is therefore injective and, since both groups have \(35\) elements, surjective. Each prime-order group is cyclic by Lagrange’s theorem. Generators \(x\) of \(P\) and \(y\) of \(Q\) give an element \((x,y)\) of order \(35\), because an integer power is the identity exactly when the exponent is divisible by both \(5\) and \(7\). Therefore \(G\cong C_{35}\), and \(C_{35}\) realizes the required order.
3. Every group of order \(24\) has a nontrivial proper normal subgroup.
Here \(n_3\) divides \(8\) and is \(1\) modulo \(3\). Of the divisors \(1,2,4,8\), exactly \(1\) and \(4\) satisfy that congruence. If \(n_3=1\), the unique subgroup of order \(3\) is nontrivial, proper, and normal.
Suppose \(n_3=4\). Conjugation on the four Sylow \(3\)-subgroups gives a homomorphism \(\rho:G\to S_4\). The action is transitive by Sylow conjugacy, so it is nontrivial and \(\ker\rho\ne G\). If \(\ker\rho\ne\{1\}\), this kernel is the required subgroup.
In the remaining case \(\rho\) is injective. Both \(G\) and \(S_4\) have order \(24\), so \(\rho\) is an isomorphism. The sign map \(S_4\to\{1,-1\}\) is onto, since a transposition has sign \(-1\). Its kernel \(A_4\) is normal of order \(12\), and \(\rho^{-1}(A_4)\) is a nontrivial proper normal subgroup of \(G\). All possibilities have been covered, so no group of order \(24\) is simple.