It's known that giving a semidirect product $(X,m)\rtimes G$ of a $G$-group $(X,m)$ with $G$ (as defined in [wiki](https://en.wikipedia.org/wiki/Semidirect_product#Outer_semidirect_products)) is the same as giving a split pair over $G$, i.e a pair of arrows $H\overset{s}{\underset{f}{\leftrightarrows}}G$ such that $s$ is a section of the lower arrow. The procedure is described e.g in [this MSE answer](https://math.stackexchange.com/a/339745/223002).




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Another approach is purely categorical, and my question is about unpacking it concretely. The category of groups is semi-abelian, whence its inverse image functor $\alpha \_G^\ast :\mathsf{Pt}\_G(\mathsf{Grp})\to \mathsf{Grp}$ is monadic. This functor is the kernel of the rightward arrow of a splitting $H\overset{s}\leftrightarrows G$. Its left adjoint takes a group $X$ to diagram below. $$X\amalg G\overset{\iota\_2}{\underset{(0\_{XG},1\_G)}{\leftrightarrows}} G$$


Thus an algebra for the monad $(T,\eta,\mu)$ induced by this adjunction is given by an arrow $$\xi:\operatorname{Ker}(0\_{XG},1\_G)\to X.$$ The object $\operatorname{Ker}(0\_{XG},1\_G)$ is comprised of trivial words in $G$ *spliced* by elements of $X$, e.g $g\_1xg\_2,g\_1x\_1g\_2x\_2g\_3x\_3\in X\amalg G$ where $g\_1g\_2=g\_1g\_2g\_3=1\in G.$ Thus, **an algebra $\xi$ seems to identify words in $\operatorname{Ker}(0\_{XG},1\_G)$ with elements of $X$.** Note $\iota\_1X\leq \operatorname{Ker}(0\_{XG},1\_G)$, so $\xi$ makes some inner identifications in $\operatorname{Ker}(0\_{XG},1\_G)$.


By monadicity, a splitting $H\leftrightarrows G$ is the same as an algebra $\xi:\operatorname{Ker}(0\_{XG},1\_G)\to X.$


* Given a splitting, the Eilenberg-Moore comparison functor endows $\operatorname{Ker}(H\overset{f}\to G)$ with the algebra map $$\alpha\_G^\ast(\varepsilon\_{H\leftrightarrows G}):\operatorname{Ker}(0\_{\operatorname{Ker}f,G},1\_G)\longrightarrow \operatorname{Ker}f.$$ If I understand correctly, $\varepsilon\_{H\leftrightarrows G}=(\ker(0\_{HG},1\_G),0\_{HG})$
so the algebra map acts by $$g\_1h\_1g\_2h\_2g\_3h\_3\mapsto h\_1h\_2h\_3,\; h\_1,h\_2,h\_3\in \operatorname{Ker}f$$ and does nothing interesting.
* Its adjoint inverse takes an algebra $(X,\xi)$ to the coequalizer of the lift of the pair $(T\xi,\mu\_X)$. This lift is the parallel pair below along with arrows to $G$ given by $(0,1\_G)$ and arrows from $G$ given by coproduct injections $\iota\_2$. $$\operatorname{Ker}(0\_{XG},1\_G)\amalg G\overset{\xi\amalg G}{\underset{\varepsilon\_{X\amalg G\leftrightarrows G}}{\rightrightarrows}} X\amalg G$$ If I understand correctly, $\varepsilon\_{X\amalg G\leftrightarrows G}=(\ker(0\_{XG},1\_G),0\_{X\amalg G,G})$. Thus the coequalizer of the above pair identifies e.g $$\xi\amalg G\;(g\_1(g\_2x\_1g\_3)g\_4(g\_5x\_2g\_6)) \sim \varepsilon\_{X\amalg G\leftrightarrows G}(g\_1(g\_2x\_1g\_3)g\_4(g\_5x\_2g\_6)) $$ i.e $$g\_1\xi(g\_2x\_1g\_3)g\_4\xi(g\_5x\_2g\_6) \sim \not \! g\_1(g\_2x\_1g\_3)\not \! g\_4(g\_5x\_2g\_6)= g\_2x\_1g\_3g\_5x\_2g\_6.$$
In particular, considering $g(x)g^{-1}\in \operatorname{Ker}(0\_{XG},1\_G)\amalg G$ with $(x)\in \operatorname{Ker}(0\_{XG},1\_G)$, we see $$gxg^{-1}=g\xi(x)g^{-1}=\xi\amalg G\;g(x)g^{-1}\sim \varepsilon\_{X\amalg G\leftrightarrows G} \;g(x)g^{-1}=x\in X\amalg G.$$ (The leftmost equality is due to the algebra axiom $\xi \circ \eta\_X=1\_X$.) This identification resembles the [usual presentation](https://math.stackexchange.com/a/160875/223002) of a semidirect product, but has no mention of an action of $G\to \mathsf{Aut}\_\mathsf{Grp}(X)$...


So, given an algebra $(X,\xi)$, the splitting associated to it is $$\mathrm{Coeq}(\xi\amalg G,(\ker(0\_{XG},1\_G),0\_{X\amalg G,G}))\leftrightarrows G$$ with the coequalizer object being called the semi-direct product $(X,\xi)\rtimes G$.




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**My questions.** In this case of groups:


1. How can I *directly* prove that giving a mysterious algebra on $X$ structure is actually the same as giving a $G$-action on $X$?
2. How can I *directly* prove that the assignments of the adjoint equivalence of the Eilenberg-Moore comparison functor are mutually inverse?



Relating three viewpoints on the semidirect product
First of all, let us see what is an algebra for this monad. One can show that the kernel of $(0,1):X\amalg G\to G$ is generated by the conjugates of elements $X$ by elements of $G$; so an action $\xi$ is in some sense a way to interpret conjugation by $G$ in $X$. To be more precise, we can define for all $g\in G$ and $x\in X$ $g\ast x=\xi (gxg^{-1})$; then the unit condition $\xi \circ\eta\_X=1\_X$ tells us that $e\_G\ast x=x$ for all $x$, while the associativity condition $\xi \circ \mu=\xi \circ T(\xi)$ tells you that $(g\_1g\_2)\ast x=g\_1\ast( g\_2\ast x)$ for all $x$, so $\ast$ is an action in the usual sense.


Now why are the comparison functor and its adjoint mutually inverse?


First of all, notice that you have the wrong counit $\varepsilon\_{H \leftrightarrows G}$: it should be a morphism
$$\left(\operatorname{Ker} f\amalg G\overset{\iota\_2}{\underset{(0\_{XG},1\_G)}{\leftrightarrows}} G\right) \rightarrow \left(H\overset{s}{\underset{f}{\leftrightarrows}}G\right)$$
in the category $\mathsf{Pt}\_G(\mathsf{Grp})$, which means in particular that $\varepsilon\_{H\leftrightarrows G}\circ \iota\_2=s$. So in fact you should have $\varepsilon\_{H\leftrightarrows G}=(\ker(0\_{HG},1\_G),s)$.


Now if you take a word of the form $g\_1hg\_2$ such that $g\_1g\_2=1$ in $G$. This condition is equivalent to $g\_2=g\_1^{-1}$, so such a word is really just the conjugate of $x\in \operatorname{Ker}f$ in $\operatorname{Ker}f\amalg G$; and now the action $\xi=\alpha\_G^\ast(\varepsilon\_{H\leftrightarrows G})$ will map $g\_1xg\_1^{-1}$ to $s(g\_1)x s(g\_1^{-1})=s(g\_1)xs(g\_1)^{-1}$. So the action $\xi=\alpha\_G^\ast(\varepsilon\_{H\leftrightarrows G})$ is simply the conjugation of $G$ on $\operatorname{Ker}f$ (through $s$), computed in $H$.


And for the left adjoint : this time the counit is given by $\varepsilon\_{X\amalg G\leftrightarrows G}=(\ker(0\_{XG},1\_G),\iota\_2)$. So what the coequalizer does is identifying words of the form $gxg^{-1}$ with $\xi(gxg^{-1})=g\ast x$. Thus the semi-direct product is generated by $X$ and $G$ with the condition that conjugation of $G$ on $X$ is given by the (classical) action $\ast$.


