The following is an exercise in Hungerford's abstract algebra text.
If $N$ and $K$ are normal subgroups of a group $G$ such that $G=MN$ and $M\cap N=\langle e \rangle$ then $G=M\times N$.
If $G=S_3$ with $N=\langle (123) \rangle,$ and $N=\langle (12)\rangle,$ with $N$ as not being normal, then the conclusion to the above statement would not hold. However, for finite permutation groups, say $G$ is a finite permutation group. If it is possible to write $G$ as an external direct product of its subgroups $P,Q$ (and assuming such can be done), then for any $\sigma\in G,$ $\sigma=(\sigma_1, \sigma_2)\in P\times Q?$ The reason I am asking is because I was never told when we take product of permutations, is that consider an external direct product. But notationally, when talking about external direct product, we use $n-$tuple notation $(a_1,a_2,...,a_n)$ instead of $a_1a_2a_3\cdots a_n$
Thank you in advance
Isomorphism: $\phi(g) = \phi(mn) = (m,n) \in M \times N$.
Multiplication in $M \times N$: $$(m_1,n_1) \times (m_2,n_2) = (m_1m_2,m_2^{-1}n_1m_2n_2)$$
$$\phi(m_1n_1m_2n_2) = \phi(m_1m_2m_2^{-1}n_1m_2 n_2) = (m_1m_2,m_2^{-1}n_1m_2 n_2)$$
$$\phi(m_1n_1) \times \phi(m_2n_2) = (m_1,n_1) \times (m_2,n_2) = (m_1m_2,m_2^{-1}n_1m_2n_2) = \phi(m_1n_1m_2n_2)$$
Hence $\phi$ is a homomorphism.
$\phi$ is injective: $\phi(m_1n_1) = (m_1,n_1) = (e,e) \implies m_1 = e, n_1 = e$
$\phi$ is clearly surjective.
Well definedness: $m_1n_1 = m_2 n_2 \implies$ $m_1 = m_2$ and $n_1 = n_2$ since $M \cap N = \{e\}$.
Hence $\phi$ is an isomorphism and $MN \approx M \times N$. But here the multiplication in $M \times N$ is defined same as semi-direct product of $M$ and $N$. So technically $MN$ is isomorphic to semi-direct product of $M$ and $N$.