Definition: Suppose $G$ is a group with normal subgroup $H$ and that $G/H\cong K$ then $G$ is an extension of $H$ by $K$
Let $\phi$ be an isomorphism of $G/H$ onto $K$. Let $X$ be a left transversal of $H$ in $G$. If $g\in G, g=xh$ for some $x\in X,h\in H$
is this expression unique? Like I try $xh_1=xh_2\implies h_1=h_2$.
Then $$gx\in \text{some coset }yH\implies gx=yh=ym_{g,x}\tag1$$ $(gH)\phi=k\in K$, let $x_k\in X$ be the representative of $gH$. $X=\{x_k:k\in K\}$ and $x_1=1$.
Now, $(x_kx_{k'}H)\phi=kk'$ and using $(1)$, $$x_kx_{k'}=x_{kk'}m_{x_k x_{k'}}=x_{kk'}m_{kk'}\quad[\text{shorthand notation}]$$ Every element $g\in G$ can be written uniquely in the form $x_k h$ where $x_k\in X,h\in H$. Product of two element, $$x_k h\cdot x_{k'} h'=x_{kk'}m_{k,k'}(x^{-1}_{k'}hx_{k'})h'=x_{kk'}m_{k,k'}h^{k'}h'$$
How they come up with this product?
The group $G$ is made up of two mappings:
- a mapping $m$ from $K\times K$ into $H$
- a mapping $\alpha$ of $K$ into a set of mappings of $H$ into $H$ (the effect of $k\alpha$ is to map $h$ to $h^{k'}$), $$k\alpha:h\rightarrow h^k$$
If enough conditions are added to these maps, one can reverse the procedure. Consider the split extension.
The second part of the question is here, which focus on split extension.
Lastly, I couldn't catch the whole story, it seems like the construction isn't intuitive enough. And I couldn't get the similarity with the Wikipedia definition,
A split extension is an extension $1\to K\to G\to H\to 1$ with a homomorphism $s\colon H\to G$ such that going from $H$ to $G$ by $s$ and then back to $H$ by the quotient map of the short exact sequence induces the identity map on $H$ i.e., $\pi \circ s={\mathrm {id}}_{H}$. In this situation, it is usually said that $s$ splits the above exact sequence.
