Relating three viewpoints on the semidirect product

Question

It's known that giving a semidirect product $(X,m)\rtimes G$ of a $G$-group $(X,m)$ with $G$ (as defined in wiki) 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.

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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 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?

Answer 1

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 $\ast$.