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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 $\ker \pi = \iota (N) = N$?

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

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