Understanding a proof about splitting of short exact sequences.

I am reading a paper by Keith Conrad about the splitting of exact sequences. I have a few questions about one particular section.

This is Theorem 3.3 in this paper http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/splittinggp.pdf.

Theorem 3.3. Let $\ \ 1\rightarrow H \overset{\alpha}{\rightarrow} G \overset{\beta}{\rightarrow} K \rightarrow 1$ be a short exact sequence. The following are equivalent.

(i) There is a homomorphism $\ \beta^{'}:K\rightarrow G \$ s.t. $\beta(\beta^{'}(k))=k$ for all $k\in K$

(ii) There is a homomorphism $\sigma:K\rightarrow \text{Aut}(H)$ and an isomorphism $\theta:G\rightarrow H\times_{\sigma}K$ s.t. the following diagram commutes.

(I will not show the diagram because my question does not concern it.)

Proof: $(i)\implies (ii)$ From the homomorphism $\beta^{'}$ we have to create an action $\sigma$ of $K$ on $H$ by automorphisms and an isomorphism of $G$ with $H\times_\sigma K$ . Using $\beta^{'}$, we can make $K$ act on $H$ using conjugation in $G$: for $k\in K$ and $h\in H$, $\beta^{'}(k)\alpha(h)\beta^{'}(k^{-1})\in \ker(\beta)$. Since $\text{img}(\alpha)=\ker(\beta)$ there exists $h'\in H$ s.t. $\alpha(h')=\beta^{'}(k)\alpha(h)\beta^{'}(k^{-1})$ and $h'$ is unique since $\alpha$ is injective. Set $\sigma_k(h)=h'$, so $\sigma_k(h)$ is the unique element in $H$ s.t.

$\alpha(\sigma_k(h))=\beta^{'}(k)\alpha(h)\beta^{'}(k)^{-1} \ \ \ \ \ \ \ \ \ \ \ (1)$

We want to show $k\mapsto \sigma_k$ is a homomorphism from $K$ to $\text{Aut}(H)$.

First, setting $k=1$ in $(1)$, $\alpha(h)=\alpha(\sigma_1(h))$ so $\sigma_1(h)=h$ for all $h$. Thus $\sigma_1=\text{id}_H$.

next we check $\sigma_k:H\rightarrow H$ is a homomorphism for each $k\in K$. For $h_1$ and $h_2$ in $H$, $\sigma_k(h_1h_2)$ is charachterized by the equation $\beta^{'}(k)\alpha(h_1h_20\beta^{'}(k)^{-1}=\alpha(\sigma_k(h_1h_2)).$ The left side is .....(it then goes on to conclude that $\sigma_k$ is a homomorphism and I understand this part).

Next we show $\sigma_{k_1}\circ \sigma_{k_2}=\sigma_{k_1k_2} \ \ \ \$ ...(it then goes on to show this is true, and I understand this, in fact until now I understand all of the computation, but not the motivation. These next few lines are where I get lost.)

In particular $\sigma_k \circ \sigma_{k^{-1}}=\sigma_1$ and $\sigma_{k^{-1}}\circ \sigma_k=\sigma_1$ so $\sigma_k\in \text{Aut}(H)$ and $k\mapsto \sigma_k$ is a homomorphism $K \rightarrow \text{AUT}(H).$ We have shown that (1) provides an action of $K$ on $H$ by automorphims so we have the semidirect product $H\times_{\sigma}K$.

The questions I have are.

(1) I understand that we showed $\sigma_k$ is a homomorphism, but we never showed injectivity or surjectivity, so did he just assume it was trivial?

(2) why did we show that $\sigma_1$ is the identity?

(3) why did we show $\sigma_{k^{-1}}\circ \sigma_k=\sigma_1$?

(4) from what I understand of semidirect products we need of the groups to be Normal (are we just considering the image of $\alpha$)?

(5) how have we shown that we have a semidirect product? and what does the action have to do with things?(I suspect these questions are related)

(6) I understand all of the computation but not the motivation or the outcome, if you could explain anything you think I could be missing out on that would be great.

I know this is long, so thank you for any help.

$\sigma_k \circ \sigma_{k^{-1}} = \sigma_{k^{-1}} \circ \sigma_k = \sigma_{1_K} = 1_H$
Only $H$ needs to be normal in $H \rtimes_{\sigma} K$.
An "external" semi-direct product (where we build the semi-direct product up from two smaller groups) is isomorphic to an "internal" semi-direct product (where we consider a parent group as a semi-direct product of two subgroups). In the "external" semi-direct product, we need the $K$-action on $H$ via $\sigma$, in the "internal" semi-direct product, this action is conjugation of $H$ by an element in $K$.