Let \(G\) be a finite group of order 84 with 28 Sylow \(3\)-subgroups. For any subgroup \(H\) of \(G\), let \(N_G(H)\) and \(Z_G(H)\) be the normalizer and the centralizer of \(H\) in \(G\), respectively.
Show that \(G\) has a normal Sylow \(7\)-subgroup. Let \(K\) denote this subgroup. Describe \(\mathrm{Aut}(K)\), the automorphism group of \(K\).
Show that \(Z_G(K)\) is a normal subgroup of \(G\), and that \(G/Z_G(K)\) is cyclic. (Note: since \(K\) is normal in \(G\), \(G\) acts on \(K\) by conjugation.)
Let \(Q\) be a Sylow \(3\)-subgroup of \(G\). Show that \(N_G(Q) = Z_G(Q) = Q\).
Show that \(|Z_G(K)|\) is not divisible by \(3\).
Prove that \(Z_G(K)\) must have order \(28\).
Solution
Write \(C=Z_G(K)\). We use \(Z_G(H)\) for the elements of \(G\) commuting with every element of \(H\).
1. The Sylow \(7\)-subgroup \(K\) is normal and\(\operatorname{Aut}(K)\cong C_6\).
Sylow’s theorems give \(n_7\mid12\) and \(n_7\equiv1\pmod7\) [@DF04]. Among \(1,2,3,4,6,12\), only \(1\) has that congruence, so \(K\) is unique and therefore normal. It has order \(7\) and is cyclic; fix a generator \(k\).
An automorphism must send \(k\) to one of the six generators \(k^j\), \(1\leq j\leq6\). Conversely each such assignment defines an automorphism. Composition multiplies exponents modulo \(7\), giving \(\operatorname{Aut}(K)\cong(\mathbb Z/7\mathbb Z)^\times\). The successive powers of \(3\) modulo \(7\) are \(3,2,6,4,5,1\), so this group is cyclic of order \(6\).
2. The subgroup \(C\) is normal and \(G/C\) is cyclic of orderdividing \(6\).
Normality of \(K\) makes conjugation a homomorphism \[ \theta:G\longrightarrow\operatorname{Aut}(K),\qquad \theta(g)(k)=gkg^{-1}. \] Its kernel consists exactly of the elements commuting with all of \(K\), so \(\ker\theta=C\). A kernel is normal. The first isomorphism theorem identifies \(G/C\) with a subgroup of the cyclic group of order \(6\) found in step 1 [@DF04].
3. For every Sylow \(3\)-subgroup \(Q\),\(N_G(Q)=Z_G(Q)=Q\).
Conjugation is transitive on the Sylow \(3\)-subgroups, and the stabilizer of \(Q\) is \(N_G(Q)\). Thus \([G:N_G(Q)]=n_3=28\), giving \(|N_G(Q)|=84/28=3\). Since \(Q\subseteq N_G(Q)\) and \(|Q|=3\), the normalizer is \(Q\). The cyclic group \(Q\) centralizes itself, while any element centralizing \(Q\) normalizes it. Therefore \(Q\subseteq Z_G(Q)\subseteq N_G(Q)=Q\).
4. The integer \(|C|\) is not divisible by \(3\).
Otherwise Cauchy’s theorem would give \(q\in C\) of order \(3\) [@DF04]. Its subgroup \(Q=\langle q\rangle\) would be Sylow in \(G\). Since \(q\) commutes with every element of \(K\), so does each power of \(q\), and hence \(K\subseteq Z_G(Q)\). Step 3 would then give \(K\subseteq Q\), impossible for groups of orders \(7\) and \(3\).
5. The centralizer \(C\) has order \(28\).
Put \(d=[G:C]\). Step 2 gives \(d\mid6\), and \(84=d|C|\). Step 4 forces \(3\mid d\), so \(d=3\) or \(6\). Consequently \(|C|=28\) or \(14\).
Suppose \(|C|=14\). The abelian subgroup \(K\) lies in \(C\), and the definition of \(C\) gives \(K\subseteq Z(C)\). The quotient \(C/K\) has order \(2\) and is cyclic. Choose \(z\in C\) whose coset generates it. Every element of \(C\) is \(kz^j\) for some \(k\in K\) and integer \(j\). Since \(K\) is central, any two such elements commute. Thus \(C\) is abelian.
In this abelian group, all Sylow \(2\)-subgroups are normal, so Sylow conjugacy implies there is exactly one. It has order \(2\); write it as \(\{1,t\}\). The element \(t\) is the unique element of order \(2\) in \(C\). Since \(C\lhd G\), conjugation by any \(g\in G\) permutes the elements of \(C\) preserving their orders. It must therefore fix \(t\). Thus \(t\in Z(G)\).
3. An element of order \(2\) cannotbelong to a group of order \(3\). This contradiction excludes\(|C|=14\), leaving \(|C|=28\) as required.