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In Dummit & Foote's Abstract Algebra (2nd ed.), Corollary 20(2) states the following:

$(\mathbb{Z}/2^n\mathbb{Z})^\times$ is the direct product of a cyclic group of order 2 and a cyclic group of order $2^{n-2}$, for all $n\geq2$.

For the proof, it says "it directly follows from Exercises 22 and 23 of Section 2.3." From these exercises, I learned that $5$ is an element of order $2^{n-2}$ in $(\mathbb{Z}/2^n\mathbb{Z})^\times$ for $n\geq2$, and that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ is not cyclic for $n\geq3$. But I don't see how these imply that $(\mathbb{Z}/2^n\mathbb{Z})^\times$ is a product of two cyclic groups of order 2 and $2^{n-2}$.

ashpool
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1 Answers1

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$|(\Bbb{Z}/2^n\Bbb{Z})^\times| = 2^{n-1}$ and index 2 subgroups are normal...

Eric Towers
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  • Does an index 2 (normal) subgroup necessarily split the group into a direct product? I don't think so... consider a semidirect product $G\rtimes\mathbb{Z}/2\mathbb{Z}$, for example... – ashpool May 31 '20 at 05:28
  • @ashpool : You only ask for "a product". – Eric Towers Jun 02 '20 at 17:53