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...

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.