If $M, N$ are normal subgroups in a group $G$ with $M \cap N = 1$, then $mn =nm $ for all $m\in M $ and $n\in N $, and also show that if $MN =G $ then $G \cong M * N $.
I would really appreciate it if someone could show me a rigorous proof for both of these. Shouldn't the first one be obvious from the definition of normal subgroup? As for the second, how do we draw out the isomorphism?
A nice way to show this is by means of commutators: for subsets $\;A,B\;$ of a group $\;G\;$ , define
$$[A,B]=\langle\,[a,b]:=a^{-1}b^{-1}ab\;|\;a\in A,\,b\in B\,\rangle$$
Be sure you can prove that $\;N\lhd G\iff [A,N]\le N\;$ for all $\;A\subset G\;%$ , and thus we have that
$$[N,M]\le N\cap M=1\iff\forall\,n\in N,\,m\in M,\;\;1= [n,m]=n^{-1}m^{-1}nm\iff nm=mn$$