"Short exact sequence split" answer illustration. Here is the question from this link Short exact sequence split
For groups $G$, $H$, and $K$, assume there exists a left-split short exact sequence: $$ 1 \rightarrow K \xrightarrow{\varphi} G \xrightarrow{\psi} H \rightarrow 1$$
Then $\varphi$ is an injective homomorphsim, $\psi$ is a surjective homomorphism, and ${\rm Im}(\varphi) = \ker(\psi)$. Furthermore, there exists a homomorphism $\pi: G \rightarrow K$ such that $\pi \circ \varphi = id_K$.
How can I show that these assumptions imply that
$H \triangleleft G, K \triangleleft G, G = HK,$ and $H \cap K = \{ 1 \}$?
And here is the solution from the same link:
Before we begin, I would like to set forth the following general relations which we shall refer to in the course of our proof:

*

*For any group morphism $f \colon G \to G'$ and any subset $X \subseteq G$, we have that $f^{-1}\left[f[X]\right]=X\mathrm{Ker}f$.

*For any group morphism $f \colon G \to G'$ and any subgroup $H \leqslant G$, we have the following description for the kernel of the restriction: $\mathrm{Ker}f_{|H}=H \cap \mathrm{Ker}f$.

For convenience I will slightly alter the original notation. Consider the following exact sequence:
$$\{1\} \xrightarrow \ F \xrightarrow{f} E \xrightarrow{g} G \xrightarrow{} \{1\} \tag{*}
$$
of groups, where $f$ admits the retraction $h \colon E \to F$. Consider the subgroups $H\colon=\mathrm{Im}f=\mathrm{Ker}g \trianglelefteq E$ and $K\colon=\mathrm{Ker}h \trianglelefteq E$.
The relation $h \circ f=\mathbf{1}_F$ leads to $h[H]=F$, whence by taking inverse images through $h$ we derive $E=h^{-1}[F]=h^{-1}\left[h[H]\right]=HK$ (general relation 1).
Since by definition $\mathrm{Im}f \subseteq H$ we have $\mathbf{1}_F=h \circ f=h_{|H} \circ {}_{H|}f$ (for arbitrary map $k \colon A \to B$ with subsets $M \subseteq A$, $N \subseteq B$ such that $k[M] \subseteq N$, the symbol ${}_{N|}k_{|M}$ denotes the restriction of $k$ between $M$ and $N$). Being the restriction of a map to its image, ${}_{H|}f$ is surjective and since it is the restriction of an injection it continues to be injective. This means that ${}_{H|}f$ is an isomorphism and the previous relation entails that the restriction $h_{|H}=\left({}_{H|}f\right)^{-1}$ is the inverse isomorphism. In particular this means that $h_{|H}$ is injective and we thus have $\{1_E\}=\mathrm{Ker}h_{|H}=K \cap H$ (general relation 2).
At this point we have already established that $H$ and $K$ are mutually supplementary subgroups of $E$, hence $E \approx H \times K \hspace{3pt} (\mathbf{Gr})$. Since ${}_{H|}f$ is an isomorphism it is clear that $F \approx H \hspace{3pt} (\mathbf{Gr})$. Let us also inspect the relation between $K$ and $G$. As $g$ is surjective we have $G=g[E]=g[HK]=g[K]$ ($H$ being the kernel of $g$). Furthermore, $\mathrm{Ker}g_{|K}=H \cap K=\{1_E\}$, which means that the restriction $g_{|K}$ is an isomorphism as well and we thus have $K \approx G \hspace{3pt} (\mathbf{Gr})$.
The previous analysis shows that $E \approx F \times G \hspace{3pt} (\mathbf{Gr})$. Let us remark that given the context there is an explicit way of exhibiting an isomorphism not only between the forementioned groups, but actually between the extensions $(^*)$ given at the beginning and the one below:
$$\{1\} \xrightarrow{} F \xrightarrow{\iota} F \times G \xrightarrow{p} G \xrightarrow{} \{1\},$$
where $\iota$ is the canonical injection given by $\iota(x)=(x, 1_G)$ and $p$ the canonical projection onto the second factor. Let us consider the direct product in restricted sense (also known as the diagonal product) $\varphi\colon=h\underline{\times}g \in \mathrm{Hom}_{\mathbf{Gr}}(E, F \times G)$. It is straightforward to see that:

*

*$\varphi \circ f=(h \circ f) \underline{\times} (g \circ f)=\mathbf{1}_F \underline{\times} \mathbf{0}_{GF}=\iota$ (for arbitrary groups $G$ and $G'$ the symbol $\mathbf{0}_{G'G}$ denotes the null morphism from $G$ to $G'$, as the category of groups does indeed have null objects)

*$p \circ \varphi=g$ by definition of direct products in restricted sense.

This establishes the commutativity of the following diagram:

which means nothing else than that $\varphi$ is indeed a morphism of extensions, hence implicitly an isomorphism between $E$ and $F \times G$.
My questions are:
1- I do not understand from where this line in the answer is correct "(recall that in general $f^{-1}[f[X]]=X\operatorname{Ker}f$ for any group morphism $f \colon G \to G'$ and any subset $X \subseteq G$)." Could anyone please clarify that?
2- $H$ is not a subset of $F$ in our case here so how can I intersect it with $\operatorname{Ker}f$?
3- I do not understand this statement "hence $E \approx H \times K \hspace{3pt} (\mathbf{Gr})$."is correct, could anyone explain this for me, please?
4- I do not understand this statement also "$g[HK]=g[K]$ ($H$ being the kernel of $g$)." why $H$ being $\operatorname{Ker}g$ makes us do so?
5- I do not understand also this statement "Furthermore, $\operatorname{Ker}g_{|K}=H \cap K=\{1_E\}$, which means that the restriction $g_{|K}$ is an isomorphism as well", why the intersection equals $\{1_E\}$? and why that means that $g_{|K}$ is an isomorphism, could anyone explain that for me, please?
6- I do not understand how "The previous analysis shows that $E \approx F \times G \hspace{3pt} (\mathbf{Gr})$." could anyone explain this to me please?
 A: Let me answer all your questions in order:

*

*It might not have been so at the time when you created this new posting, but I eventually modified the form of my original answer so as to contain two initial propositions, statements that are generally valid in group theory (I also took the liberty to edit this new posting so as to include those general statements). One of them is:


Proposition 1. For any group morphism $f \colon G \to G'$ and any subset $X \subseteq G$, the relation $f^{-1}[f[X]]=X\mathrm{Ker}f$ is valid.

Proof. This can be formulated for algebraic structures more general than groups, but the idea is that the multiplicative operation "$\cdot$" implicit on $G$ can be naturally extended to the powerset $\mathscr{P}(G)$ in the following manner:
$$\begin{align}
\cdot \colon \mathscr{P}(G) &\to \mathscr{P}(G)\\
\cdot(X, Y) \colon&=XY\colon=\{xy\}_{\substack{x \in X\\y \in Y}}.
\end{align}$$
It is easy to ascertain that the newly defined structure $(\mathscr{P}(G), \cdot)$ is a monoid with unity $1_{\mathscr{P}(G)}=\{1_G\}$ (I invite you to undertake the verification as an exercise, it should prove to be a simple, enjoyable activity). Furthermore, since $f$ is a morphism and thus "commutes" with products of elements it will also "commute" with products of subsets in the sense that $f[XY]=f[X]f[Y]$ for any subsets $X, Y \subseteq G$ (the product on the right-hand side of this equality is of course considered in the analogous monoid $\mathscr{P}\left(G'\right)$). In an even more formal parlance, the map:
$$\begin{align}
\widehat{f} \colon \mathscr{P}(G) &\to \mathscr{P}\left(G'\right)\\
\widehat{f}(X)\colon&=f[X]=\{f(x)\}_{x \in X}
\end{align}$$
obtained by extending $f$ between the powersets is actually a monoid morphism, $\widehat{f} \in \mathrm{Hom}_{\mathbf{Mon}}\left(\mathscr{P}(G), \mathscr{P}\left(G'\right)\right)$.
Let us also note that given any nonempty subset $\varnothing \neq X \subseteq \mathrm{Ker}f$ we have $f[X]=\{1_{G'}\}$. Indeed, since $X \neq \varnothing$ it follows that $f[X] \neq \varnothing$ and from the definition of the kernel we have $f[X] \subseteq \left\{1_{G'}\right\}$. Since the only nonempty subset of a singleton is itself, the desired conclusion follows. Since the kernel itself is a subgroup and therefore nonempty, this applies in particular to $X=\mathrm{Ker}f$ ($\color{red}{this}$ also directly relates to question 4). Thus, it is clear that $f\left[X\mathrm{Ker}f\right]=f[X]f[\mathrm{Ker}f]=f[X]\{1_{G'}\}=f[X]$, which means that $X\mathrm{Ker}f \subseteq f^{-1}\left[f[X]\right]$.
As to the reverse inclusion, consider an arbitrary $y \in f^{-1}\left[f[X]\right]$. This means there exists $x \in X$ such that $f(y)=f(x)$ and therefore that $f(x)^{-1}f(y)=f\left(x^{-1}y\right)=1_{G'}$, which further entails $x^{-1}y \in \mathrm{Ker}f$. We thus have $y=x\left(x^{-1}y\right) \in X\mathrm{Ker}f$ and by the arbitrariness of $y$ conclude that $f^{-1}\left[f[X]\right] \subseteq X\mathrm{Ker}f$. $\Box$


*In the original formulation, the $H$ and $f$ mentioned in the quoted statement are not the same $H$ and $f$ as those in the exact sequence problem (I merely ran out of symbols I preferentially use to denote such objects and ended up repeating the same notation but with different meaning, hence the confusion). The confusion I hope is clarified, now that I stated the general propositions needed for the argument on a separate level, before starting the argument itself.

*Let us recall the setting in which the quoted statement was made:


Proposition 2. Let $E$ be a group possessing normal subgroups $H, K \trianglelefteq E$ such that $E=HK$ and $H \cap K=\{1_E\}$. Then we have the group isomorphism $E \approx H \times K$.

Proof. Let us consider the map:
$$\begin{align}
\varphi: H \times K &\to E\\
\varphi(x, y)&=xy
\end{align}$$
and let us argue that it is a group morphism. In order to show this, it will suffice to prove that any element of $H$ commutes with any element of $K$, which in a more succinct formulation can be expressed as $H \leqslant \mathrm{C}_G(K)$ (the latter object is the centraliser of $K$ in $G$). Consider thus arbitrary $x \in H$ and $y \in K$ together with their commutator $[x, y]=(yx)^{-1}xy=x^{-1}y^{-1}xy$. We have on the one hand $[x, y]=\left(xy^{-1}x\right)y \in KK=K$ -- since $xy^{-1}x^{-1}$ is a conjugate of the element $y^{-1}$ of the normal subgroup $K$ -- and on the other hand $[x, y]=x^{-1}\left(y^{-1}xy\right) \in HH=H$, since $y^{-1}xy$ is a conjugate of the element $x$ of the normal subgroup $H$. We thus derive $[x, y] \in H \cap K=\{1_E\}$, which means by definition of commutators that $xy=yx$, Q.E.D.
The above justifies the fact that $\varphi \in \mathrm{Hom}_{\mathbf{Gr}}(H \times K, E)$. It is clear by definition that $\mathrm{Im}\varphi=HK$, so the hypothesis $HK=E$ entails the surjectivity of $\varphi$. It is equally clear that $\mathrm{Ker}\varphi=\left\{\left(t, t^{-1}\right)\right\}_{t \in H \cap K}$, whence from the hypothesis $H \cap K=\{1_E\}$ of trivial intersection we gather that $\varphi$ has trivial kernel and is thus injective. Combining all these observations, we deduce that $\varphi$ is an isomorphism. $\Box$


*This is a direct consequence of $\color{red}{what\ was\ discussed}$ in the proof of proposition 1.

*By virtue of the second general proposition at the very beginning of the edited version of my original answer, we have the description $\mathrm{Ker}g_{|K}=\mathrm{Ker}g \cap K=H \cap K$. The fact that this latter intersection $H \cap K=\{1_E\}$ is trivial was proved at an earlier step of the argument presented in the original answer. The fact that the restriction $g_{|K}$ has trivial kernel means it is injective. As to its surjectivity, the statement referenced in your previous question number 4 – which I hope is by now clarified – means that $G=g[K]$. It is clear from elementary set theory that $\mathrm{Im}g_{|K}=g_{|K}[K]=g[K]=G$, so that the restriction $g_{|K}$ is also seen to be surjective. Being simultaneously injective and surjective, $g_{|K}$ is bijective and we know that in the case of groups the notions of “isomorphism” and “bijective morphism” coincide (comforting phenomenon, which does not hold in other categories, such as that of topological spaces or of graphs).

*Once we have justified the group isomorphisms $F \approx H \hspace{3pt} (\mathbf{Gr})$ (via the isomorphism ${}_{H|}f$) and $K \approx G \hspace{3pt} (\mathbf{Gr})$ (via the isomorphism $g_{|K}$), we appeal to another very general proposition, namely that direct products of isomorphic groups remain isomorphic groups: given families $\Gamma$ and $\Gamma’$ of groups indexed by the same index set $I$, if the isomorphism relation $\Gamma_i \approx \Gamma’_i \hspace{3pt} (\mathbf{Gr})$ occurs for every index $i \in I$, then we also have the isomorphism $\displaystyle\prod_{i \in I}\Gamma_i \approx \displaystyle\prod_{i \in I}\Gamma’_i \hspace{3pt} (\mathbf{Gr})$. If $\gamma \in \displaystyle\prod_{i \in I}\mathrm{Iso}_{\mathbf{Gr}}(\Gamma_i, \Gamma’_i)$ is a family of isomorphisms, then the direct product $\eta$ of family $\gamma$ -- i.e. the unique morphism $\theta$ such that $\pi'_i \circ \theta=\gamma_i \circ \pi_i$ for every index $i \in I$, $\pi_i$ and $\pi'_i$ being the respective canonical projections of the direct products of families $\Gamma$ respectively $\Gamma'$ -- is also an isomorphism between the direct products.

