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This question already has an answer here:

I need some help with the following question.

Show that $(\mathbb{Z}/2^n \mathbb{Z})^{\times}$ is not cyclic for any $n > 3.$

There is the following hint in my book: find two distinct subgroups of order $2.$

But I don't see any subgroups of order $2.$ Is group $(\mathbb{Z}/2 \mathbb{Z})^{\times}$ subgroup of $(\mathbb{Z}/2^n \mathbb{Z})^{\times}?$ Show the second subgroup, please.

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marked as duplicate by Travis, Johanna, graydad, user147263, PVAL-inactive Mar 7 '15 at 2:50

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    $\begingroup$ Remember the superscript $\times$ means the multiplicative group of invertible elements in the ring $\mathbb{Z}/2^n\mathbb{Z}$. Here $(\mathbb{Z}/2\mathbb{Z})^\times$ is not of order $2$ (nor is it quite clear how you make that a subgroup of $(\mathbb{Z}/2^n\mathbb{Z})^\times$ for $n \gt 1$, except in the sense that it corresponds to the trivial subgroup). $\endgroup$ – hardmath Mar 6 '15 at 23:46
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Since $(2^n-1)^2 \equiv 1 \pmod {2^n}$, $\langle 2^{n}-1 \rangle$ is a subgroup of order 2. Also, since $n \geq 3$, $(1+2^{n-1})^2 \equiv 1 \pmod{2^n}$, so $\langle 1+2^{n-1} \rangle$ has order 2. It should be easy to see that these subgroups are distinct, as $n \neq 1$.

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