# Equivalent Conditions of Split Extension of Groups

Definition of split extension of $$Q$$ by $$N$$: An extension of $$Q$$ by $$N$$,

$$1 \longrightarrow N \longrightarrow G \longrightarrow Q \longrightarrow 1,$$

is said to be split if it is isomorphic to the extension defined by a semidirect product $$N \rtimes_\theta Q$$.

Does this mean that

$$1 \longrightarrow N \longrightarrow G \longrightarrow Q \longrightarrow 1$$

is parallel to

$$1 \longrightarrow N \longrightarrow N \rtimes_\theta Q \longrightarrow Q \longrightarrow 1$$

for some $$\theta : Q \to \operatorname{Aut} (N)$$, and that everything commutes?

Do we know what the maps involved in either diagram look like, or do we merely know that that there are injections from $$N \to G$$ and from $$N \to N \rtimes_\theta Q$$, and surjections from $$G \to Q$$ and from $$N \rtimes_\theta Q \to Q$$?

Also, there is an isomorphism connecting $$G$$ and $$N \rtimes_\theta Q$$, right?

I am trying to show that an extension of $$Q$$ by $$N$$ being split is equivalent to the following two statements:

$$(1)$$ there exists a subgroup $$Q ' \le G$$ such that $$\pi$$ induces an isomorphism $$Q' \to Q$$;

$$(2)$$ there exists a homomorphism $$s : Q \to G$$ such that $$\pi \circ s = id$$.

I spent most of this past week trying to show the definition is equivalent to conditions are equivalent, and I still can't even figure out how to show that the definition implies $$(1)$$. I could use some help...

• You are missing some detail here - like what is $\pi$? You ask if there is an isomorphism connecting $G$ and $N\rtimes_\theta Q$, but is this not the assumption? – Robert Chamberlain Feb 24 at 14:27
• @RobertChamberlain $\pi$ is a surjective homomorphism from $G$ to $Q$. But if two sequences are isomorphic, doesn't that mean their middle terms are isomorphic as groups? – user193319 Feb 24 at 14:43

A short exact sequence $$1 \longrightarrow N \longrightarrow G \longrightarrow Q \longrightarrow 1$$ splits if and only if $$G$$ is the outer semi direct product of $$Q$$ by $$N$$, if and only if there is such a section $$s: Q\rightarrow G$$. I just gave the proofs in my lecture, Proposition $$1.1.24$$ and $$1.1.25$$. About equivalent extension, see the section afterwards.