Let \(G\) be a group of order 84 with 28 Sylow 3-subgroups.
Find the number of Sylow 7-subgroups of \(G\).
Let \(Q\) be a Sylow 3-subgroup of \(G\). Let \(N_G(Q)\) be the normalizer of \(Q\) in \(G\), and \(Z_G(Q)\) be the centralizer of \(Q\) in \(G\). Show that \(N_G(Q) = Z_G(Q) = Q\).
Solution
1. There is exactly one Sylow \(7\)-subgroup.
Since \(84=2^2\cdot3\cdot7\), Sylow’s theorems give \[ n_7\mid12,\qquad n_7\equiv1\pmod7 \] for the number \(n_7\) of Sylow \(7\)-subgroups [@DF04]. The positive divisors of \(12\) are \(1,2,3,4,6,12\), and only \(1\) is congruent to \(1\) modulo \(7\). Thus \(n_7=1\).
2. The normalizer of \(Q\) has order \(3\), and equals \(Q\).
Conjugation acts transitively on the Sylow \(3\)-subgroups, by Sylow conjugacy [@DF04]. The stabilizer of \(Q\) is \[ N_G(Q)=\{g\in G:gQg^{-1}=Q\}. \] More explicitly, the map \(gN_G(Q)\mapsto gQg^{-1}\) is a bijection from its left cosets to the Sylow \(3\)-subgroups: two conjugates coincide exactly when the representatives differ by an element of \(N_G(Q)\). Hence \[ [G:N_G(Q)]=28,\qquad |N_G(Q)|=84/28=3. \] Every subgroup normalizes itself, so \(Q\subseteq N_G(Q)\). As a Sylow \(3\)-subgroup, \(Q\) also has order \(3\). The containment is therefore equality.
3. The centralizer is also \(Q\).
Here \(Z_G(Q)\) denotes the set of elements commuting with every element of \(Q\). The group \(Q\) has prime order \(3\) and is cyclic, hence abelian. Thus \(Q\subseteq Z_G(Q)\). Every element centralizing \(Q\) fixes each of its elements under conjugation and therefore normalizes \(Q\). Combining this with step 2 gives \[ Q\subseteq Z_G(Q)\subseteq N_G(Q)=Q. \] All three subgroups coincide, as required.