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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...

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  • $\begingroup$ 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? $\endgroup$ – Robert Chamberlain Feb 24 at 14:27
  • $\begingroup$ @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? $\endgroup$ – user193319 Feb 24 at 14:43
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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.

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