Let \(H,K\le G\). Assume \[ G=HK, \qquad H\cap K=\{e\}, \qquad H\trianglelefteq G, \qquad K\trianglelefteq G. \] Prove that the multiplication map \[ \Gamma:H\times K\longrightarrow G, \qquad \Gamma(h,k)=hk, \] is an isomorphism. Determine how the normality hypotheses can be weakened.
Solution
1. Every element of \(H\) commutes with every element of \(K\).
Let \(h\in H\) and \(k\in K\). Because \(H\trianglelefteq G\), \[ kh^{-1}k^{-1}\in H, \] so \[ [h,k]=hkh^{-1}k^{-1}=h(kh^{-1}k^{-1})\in H. \] Because \(K\trianglelefteq G\), \[ hkh^{-1}\in K, \] so \[ [h,k]=(hkh^{-1})k^{-1}\in K. \] Thus \[ [h,k]\in H\cap K=\{e\}, \] hence \([h,k]=e\) and therefore \(hk=kh\).
2. The map \(\Gamma\) is a homomorphism.
For \((h_1,k_1),(h_2,k_2)\in H\times K\), 1 gives \(k_1h_2=h_2k_1\). Hence \[ \begin{aligned} \Gamma(h_1,k_1)\Gamma(h_2,k_2) &=h_1k_1h_2k_2\\ &=h_1h_2k_1k_2\\ &=\Gamma(h_1h_2,k_1k_2). \end{aligned} \]
3. The map \(\Gamma\) is surjective.
The hypothesis \(G=HK\) says exactly that every \(g\in G\) has the form \(g=hk\) with \(h\in H\), \(k\in K\). Thus every \(g\) lies in the image of \(\Gamma\).
4. The map \(\Gamma\) is injective.
Suppose \[ \Gamma(h,k)=hk=e. \] Then \[ h=k^{-1}. \] The left side belongs to \(H\) and the right side belongs to \(K\), so \[ h=k^{-1}\in H\cap K=\{e\}. \] Hence \(h=e\) and \(k=e\), so the kernel is trivial. Since \(\Gamma\) is a homomorphism, it is injective.
5. Therefore\[ G\cong H\times K. \]
By <1>2–<1>4, \(\Gamma\) is a bijective homomorphism.
6. The normality assumptions can be weakened to the single condition\[ [H,K]=1, \]i.e. every element of \(H\) commutes with every element of \(K\).
The proof of 2 uses only elementwise commutation, while 3 and 4 use only \(G=HK\) and \(H\cap K=\{e\}\). Thus the three conditions \[ G=HK, \qquad H\cap K=\{e\}, \qquad [H,K]=1 \] already imply \(G\cong H\times K\).
Conversely, if the multiplication map \(H\times K\to G\) is a homomorphism, then \[ (h,e)(e,k)=(h,k)=(e,k)(h,e) \] in \(H\times K\), so their images satisfy \(hk=kh\) in \(G\). Hence elementwise commutation is exactly the condition needed in place of normality for this multiplication-map proof.
7. In the finite coprime-order case, trivial intersection is automatic.
If \(H\) and \(K\) are finite and \(\gcd(|H|,|K|)=1\), then \(|H\cap K|\) divides both \(|H|\) and \(|K|\) by Lagrange’s theorem. Hence \(|H\cap K|=1\), so \(H\cap K=\{e\}\).