Suppose we have the normal subgroups $H,J\subset G $ with the property $|G|=|H|\cdot |J|$ and $H\cap J={e}. $ $ $Prove that $H\times J\cong G$.
I don't really know how to approach this one. I thought about that $HJ=G$ and hence $HJ\cong H\times J$. But how do I construct the isomorphism?
When trying to construct isomorphism start with the most natural looking map. $$ (h,j)\mapsto hj $$ Let's call the map above $g$. Show that it is indeed a homomorphism. $\ker g=\{(h,j):hj=e\}=\{(h,j)\in H\times J:h=j^{-1}\in J\cap H=\{e\}\}=\{(e,e)\}$.