# Condition for pullback to split

This is probably an elementary question, but I'm new to this machinery.

Let $$G$$ be a group and $$N$$ be a normal subgroup of $$G$$. Let $$\Gamma$$ be another group. Suppose we have a homomorphism $$\phi:\Gamma \to G/N$$. Now we have the homomorphism $$\phi:\Gamma \to G/N$$ and the canonical quotient homomorphism $$\psi:G \to G/N$$.

Denote by $$H$$ the pullback of the two homomorphisms to $$G/N$$. That is, $$H \subseteq G \times \Gamma$$ comprising elements $$(g,\gamma)$$ such that $$\phi(\gamma)=\psi(g) \in G/N$$.

So we have the following two diagrams: $$1 \rightarrow N \rightarrow G\stackrel{\psi}{\rightarrow} G/N\rightarrow 1$$ which is a short exact sequence, and another diagram $$\begin{array} GG & \stackrel{\psi}{\longrightarrow} & G/N\\ \uparrow & & \uparrow{\phi} \\ H & \longrightarrow & \Gamma \end{array}$$ Is the following also a commutative diagram? $$\begin{array} 11 & \longrightarrow & N & \longrightarrow & G & \stackrel{\psi}{\longrightarrow} & G/N & \longrightarrow & 1\\ & & \uparrow & & \uparrow & & \uparrow{\phi} & & \\ 1 & \longrightarrow & N & \longrightarrow & H & \longrightarrow & \Gamma & \longrightarrow & 1\\ \end{array}$$ That is, can $$N$$ be embedded in the pullback $$H$$?

Secondly, suppose the map $$\phi:\Gamma \to G/N$$ can be lifted to a map $$\phi':\Gamma \to G$$ so that we have a commuting diagram bypassing $$H$$. In this case, is $$H$$ a semidirect product of $$N$$ and $$\Gamma$$?

In other words, is the bottom exact sequence split when $$\phi$$ can be lifted diagonally to $$G$$, and is the converse true too?

Thanks!

Yes and Yes. The subgroup $$N$$ of $$H$$ that you are looking for is $$\{(n,1) : n \in N \}$$, and the complement you are looking for is $$\{(\phi'(\gamma),\gamma) : \gamma \in \Gamma\}$$.
• Hello, thanks for the answer. But if you have time, could you please elaborate more? I still can't see why the existence of the lift in the diagram forces $H$ to be a semi-direct product. What is the action of $\Gamma$ on $N$ for instance? Thanks! – BharatRam Nov 30 '18 at 17:15
• I am not sure what there is to elaborate. I have given a definition of a specific complement of $N$ in $\Gamma$. The existence of a complement is equivalent to the extension being a semidirect product. – Derek Holt Nov 30 '18 at 17:34