Show that if \(H, K \normal G\) and \(H\intersect K = \{e\}\), then \(hk=kh\) for all \(h\in H,k\in K\).
Show that if \(H,K\normal G\) are normal subgroups that intersect trivially, then \([H, K] = 1\) (so \(hk = kh\) for all \(k\) and \(h\)).
Solution
1. For \(h\in H\) and \(k\in K\), the commutator \([h,k]=hkh^{-1}k^{-1}\) lies in \(H\cap K\).
Because \(H\trianglelefteq G\), conjugating \(h^{-1}\) by \(k\) gives \(kh^{-1}k^{-1}\in H\), so \[ [h,k]=h(kh^{-1}k^{-1})\in H. \] Because \(K\trianglelefteq G\), conjugating \(k\) by \(h\) gives \(hkh^{-1}\in K\), so \[ [h,k]=(hkh^{-1})k^{-1}\in K. \] Thus \([h,k]\in H\cap K\).
2. If \(H\cap K=\{e\}\), then \([h,k]=e\) for all \(h\in H\) and \(k\in K\).
By 1, every commutator \([h,k]\) lies in \(H\cap K=\{e\}\). Hence \([h,k]=e\).
3. Therefore \(hk=kh\) for all \(h\in H\) and \(k\in K\), equivalently \([H,K]=1\).
The equality \([h,k]=e\) is equivalent to \(hk=kh\). Since this holds for every pair, the subgroup generated by all commutators is trivial.