# Split Sequences. What is the Group?

See page 49 of this book.

Proposition 3.22 An extension (17) splits if $$N$$ is complete. In fact, $$G$$ is then direct product of $$N$$ with the centralizer of $$N$$ in $$G$$.

I believe (17) refers to

$$1 \rightarrow N \rightarrow Q \rightarrow Q/N \rightarrow 1,$$

where $$N$$ is a subgroup of $$Q$$ of order $$4$$.

How can we speak of the $$N$$ in $$G$$ if I don't know what $$G$$ looks like? And why does $$Q = C_G(N)$$? How do I know that $$C_G(N)$$ is the quaternion group? And why does $$G = NQ$$? I thought $$G$$ being the direct product of $$N$$ and $$Q$$ meant $$G = N \times Q$$? It is as though the author is treating $$Q$$ and $$N$$ as subgroups of $$G$$ (whatever $$G$$ is suppose to be).

EDIT (1): Perhaps the author mistakenly wrote a reference to (17). Perhaps he meant to reference (16), which is

$$1 \rightarrow N \overset{\iota}{\rightarrow} G \overset{\pi}{\rightarrow} Q \rightarrow 1$$

EDIT (2): Okay. So is this what the proposition is actually saying: If $$N$$ is a normal subgroup of $$G$$ that is complete, then exact sequence

$$1 \rightarrow N \overset{\iota}{\rightarrow} G \overset{\pi}{\rightarrow} C_G(N) \rightarrow 1$$

is a split sequence, where $$\iota : N \to G$$ is $$\iota (n) = n$$ and $$\pi : G \to C_G(N)$$ is some surjective homomorphism such that $$\ker \pi = \iota (N) = N$$?

• It's a typo. "(17)" should read "(16)". – FredH Feb 26 at 1:07
• I think you've got it right. By the way, the current version of Milne's notes has the typo corrected. – FredH Feb 26 at 1:58