Show that if every Sylow \(p\)-subgroup of a finite group \(G\) is normal for every prime \(p\), then \(G\) is the direct product of its Sylow subgroups.
Solution
Let \[ |G|=p_1^{a_1}\cdots p_r^{a_r}, \] and let \(P_i\) be the Sylow \(p_i\)-subgroup. By hypothesis each \(P_i\) is normal.
1. If \(i\ne j\), then \(P_i\cap P_j=\{e\}\).
The order of \(P_i\cap P_j\) divides both \(|P_i|=p_i^{a_i}\) and \(|P_j|=p_j^{a_j}\). Since \(p_i\ne p_j\), these orders are coprime, so \(|P_i\cap P_j|=1\).
2. If \(i\ne j\), then every element of \(P_i\) commutes with every element of\(P_j\).
Let \(x\in P_i\) and \(y\in P_j\). Because \(P_i\trianglelefteq G\), \[ [x,y]=xyx^{-1}y^{-1}\in P_i. \] Because \(P_j\trianglelefteq G\), the same commutator lies in \(P_j\). Hence \[ [x,y]\in P_i\cap P_j=\{e\} \] by 1, so \(xy=yx\).
3. The multiplication map\[ \mu:P_1\times\cdots\times P_r\longrightarrow G, \qquad (x_1,\ldots,x_r)\longmapsto x_1\cdots x_r \]is a homomorphism.
By 2, elements belonging to distinct factors commute. Therefore componentwise multiplication in the direct product is carried by \(\mu\) to multiplication in \(G\).
4. The homomorphism \(\mu\) is injective.
Suppose \(x_1\cdots x_r=e\) with \(x_i\in P_i\). Fix \(i\). Then \[ x_i=\left(\prod_{j\ne i}x_j\right)^{-1}. \] The left side has \(p_i\)-power order. The right side belongs to the product of the other Sylow subgroups, whose order divides \(\prod_{j\ne i}p_j^{a_j}\), which is coprime to \(p_i\). Hence \(x_i=e\). Since this holds for every \(i\), the kernel is trivial.
5. The homomorphism \(\mu\) is surjective.
Its domain has order \[ \prod_{i=1}^r|P_i| =\prod_{i=1}^rp_i^{a_i} =|G|. \] By 4, \(\mu\) is injective between finite groups of the same order, hence surjective.
6. Therefore\[ G\cong P_1\times\cdots\times P_r. \]
By <1>3–<1>5, \(\mu\) is an isomorphism.