Note - Internal and External Direct Products

notegroup-theory

Question

What do we mean by “internal” direct product? What is an “external” direct product?

Recall

If $(G,\cdot )$, $(H,\ast )$, we can construct an (external) direct product $G\times H$ with component-wise operations.

$$(g,h)\otimes (g^{\prime },h^{\prime })=(g\cdot g^{\prime },h\ast h^{\prime }).$$

Theorem

Given groups $(G,\cdot )$, and $N,H⊴G$ that give an internal direct product decomposition of $G$, then

$$f:H\times N\to G\text{ given by }f(h,n)=hn$$

is an isomorphism.

Proof

$N,H\le G$, $nh=hn\forall n\in N,h\in H$, $HN=G$.

$f$ is a morphism:

$$f((h,n)\otimes (h^{\prime },n^{\prime }))\stackrel{?}{=}f(h,n)f(h^{\prime },n^{\prime })$$

$f(h,n)=hn$, $f(h^{\prime },n^{\prime })=h^{\prime }{n}^{\prime }$, so $f(h,n)f(h^{\prime },n^{\prime })=hn{h}^{\prime }{n}^{\prime }=h{h}^{\prime }n{n}^{\prime }$.

$f$ is surjective: If $g\in G=HN⟹\exists h\in H,n\in N,g=hn=f(h,n)⟹f$ is onto.

$f$ is injective ($⟺\ker f=e_{H\times N}=(e_H,e_N)$) $\ker f=\{(h,n)\in H\times N:hn=f(h,n)=e_G\}$ $⟹h\in H,h=n^{-1}\in N⟹h\in H\cap N=\{e_G\}$ $⟹h=n=\{e_G\}⟹(h,n)=(e_G,e_G)=(e_H,e_N)$