Suppose \(G\) is a finite group such that every Sylow subgroup of \(G\) is normal and abelian. Show that \(G\) is abelian.
Solution
Write \[ \abs G = \prod_{i=1}^r p_i^{a_i} \] with distinct primes \(p_i\), and let \(P_i\) be a Sylow \(p_i\)-subgroup.
1. For each \(i\), the Sylow subgroup \(P_i\) is unique.
All Sylow \(p_i\)-subgroups are conjugate. Since \(P_i\) is normal by hypothesis, every conjugate of \(P_i\) equals \(P_i\). Thus there is only one.
2. If \(i\neq j\), then\[ P_i\cap P_j=\{1\}. \]
The order of \(P_i\cap P_j\) divides both \[ \abs{P_i}=p_i^{a_i} \] and \[ \abs{P_j}=p_j^{a_j}. \] These two orders are coprime, so the intersection has order \(1\).
3. If \(i\neq j\), then every element of \(P_i\) commutes with everyelement of \(P_j\).
Take \[ x\in P_i, \qquad y\in P_j. \] Since \(P_i\) is normal, \[ yx^{-1}y^{-1}\in P_i, \] so \[ [x,y] = x(yx^{-1}y^{-1}) \in P_i. \] Since \(P_j\) is normal, \[ xyx^{-1}\in P_j, \] so \[ [x,y] = (xyx^{-1})y^{-1} \in P_j. \] Step 2 therefore gives \[ [x,y]=1. \]
4. The product\[ P_1P_2\cdots P_r \]is a subgroup of \(G\) of order\[ \prod_{i=1}^r\abs{P_i} = \abs G. \]
Step 3 shows that the Sylow subgroups commute elementwise, so their setwise product is a subgroup.
Because their orders are pairwise coprime and their pairwise intersections are trivial, multiplication gives an injective homomorphism \[ P_1\times\cdots\times P_r \longrightarrow G. \] Its image is the product subgroup. Hence that subgroup has order \[ \prod_{i=1}^r p_i^{a_i} = \abs G. \]
5. One has\[ G=P_1P_2\cdots P_r. \]
Step 4 gives a subgroup of the finite group \(G\) having the same order as \(G\), so it must equal \(G\).
6. The group \(G\) is abelian.
Each \(P_i\) is abelian by hypothesis, and step 3 shows that elements from different Sylow subgroups also commute. By step 5, every element of \(G\) is a product of elements from the \(P_i\). Therefore every pair of elements of \(G\) commutes.
7. Q.E.D.
Step 6 proves the claim.