Understanding the construction of the split extension of group

Question

This is the 2nd part of another question, mainly general extension. Please have a look to understand the notation. A brief description was copied from that thread,

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 $x\in X,h\in H$
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 $(1)$, $$x_kx_{k'}=x_{kk'}m_{x_k x_{k'}}=x_{kk'}m_{kk'}\quad[\text{shorthand notation}]$$

Consider the split extension where $m_{x,x'}=1$ for all $x,x'\in X$ in $(1)\implies xx'=x''$.

Question: Can we always guarantee the existence of $m_{x,x'}=1$?

Then $X<G,H\triangleleft G$ and $G=XH$. $G$ in an extension of $H$ by $X$. Therefore we may suspect that if we are given,

• a group $H$,
• a group $K$,
• a homomorphism $\alpha$ of $K$ into the automorphism group of $H$

then we can create a splitting extension of $H$ by $K$,

$$ \begin{align} G=K\times H=&\{(k,h):k\in K,h\in H\}\\ (k,h)(k',h')=&(kk',h(k'\alpha)h')\\ (k,h)^{-1}=&(k^{-1},h^{-1}(k^{-1}\alpha)) \end{align} $$

Question: How they come up with this multiplication and the inverse?

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.

Answer 1

Your first question seems to imply that all extensions of $H$ wit $K$ split. Consider $G=C_4$ (cyclic of order $4$) with $H=K=C_2$. Regarding your second question, split extensions are more or less the same as semidirect products. You give the definition of an external semidirect product. Here is the definition of an internal semidirect product: Let $G$ be a group with $N\unlhd G$ and $H\le G$ such that $N\cap H=1$ and $G=HN$. Then $G$ is called a semidirect product of $N$ and $H$. For $h_1,h_2\in H$ and $x_1,x_2\in N$ we have $$(h_1x_1)(h_2x_2)=(h_1h_2)(h_2^{-1}x_1h_2x_2)=(h_1h_2)(c_{h_2}(x_1)x_2)$$ where $c_{h_2}$ denotes the inner automorphism induced by conjugation with $h_2$. Now you can see the similarity to your external definition.