Prove the “recognizing direct products” theorem. Can the conditions be relaxed?
Solution
The recognizing-direct-products theorem says that if \(H,K\le G\) satisfy \[ H\trianglelefteq G, \qquad K\trianglelefteq G, \qquad H\cap K=\{e\}, \qquad HK=G, \] then \(G\cong H\times K\).
1. Every element of \(H\) commutes with every element of \(K\).
Let \(h\in H\) and \(k\in K\). Since \(H\) is normal, \[ kh^{-1}k^{-1}\in H, \] so \[ [h,k]=hkh^{-1}k^{-1}\in H. \] Since \(K\) is normal, \[ hkh^{-1}\in K, \] so \([h,k]\in K\) as well. Hence \[ [h,k]\in H\cap K=\{e\}, \] and therefore \(hk=kh\).
2. The multiplication map\[ \mu:H\times K\longrightarrow G, \qquad \mu(h,k)=hk, \]is a homomorphism.
For \((h_1,k_1),(h_2,k_2)\in H\times K\), 1 gives \(k_1h_2=h_2k_1\). Thus \[ \mu(h_1,k_1)\mu(h_2,k_2) =h_1k_1h_2k_2 =h_1h_2k_1k_2 =\mu(h_1h_2,k_1k_2). \]
3. The map \(\mu\) is surjective.
The hypothesis \(HK=G\) says exactly that every \(g\in G\) can be written \(g=hk\) with \(h\in H\) and \(k\in K\).
4. The map \(\mu\) is injective.
If \(\mu(h,k)=e\), then \(hk=e\), so \[ h=k^{-1}. \] The left side lies in \(H\) and the right side lies in \(K\), hence \[ h=k^{-1}\in H\cap K=\{e\}. \] Thus \(h=k=e\), so \(\ker\mu=\{(e,e)\}\).
5. Therefore \(G\cong H\times K\).
By <1>2–<1>4, \(\mu\) is a bijective homomorphism.
6. The normality hypotheses can be replaced by the weaker-looking condition that \(H\) and \(K\) commute elementwise.
Assume only \[ HK=G, \qquad H\cap K=\{e\}, \qquad [h,k]=e\quad\text{for all }h\in H,\ k\in K. \] Then the proofs of <1>2–<1>5 apply verbatim, so \(G\cong H\times K\). Thus direct elementwise commutation is the condition actually needed to make the multiplication map a homomorphism; normality in the original theorem is a sufficient way to obtain it from the trivial-intersection hypothesis.