How can I prove that the multiplicative group $G:=\{m\in\mathbb Z_{2^n}:m\equiv 1 \mod 4\}$ is cyclic? I tried claiming that $G=\langle5\rangle$ but failed.
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This amounts to proving the existence of primitive roots – Ben Grossmann Nov 23 '14 at 19:39
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I recognized that and keep finding a primitive root, I'm stuck here. – Dave Westwood Nov 23 '14 at 19:44
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I don't understand. What do you mean by "I keep finding a primitive root"? – Ben Grossmann Nov 23 '14 at 19:45
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Trying to find one... – Dave Westwood Nov 23 '14 at 19:49
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Do you mean that you're trying to find a single integer $q$ such that we will always have $G = \langle q \rangle$, regardless of which $n$ is chosen? If so, then don't do that. – Ben Grossmann Nov 23 '14 at 19:51
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No. Finding primitive roots depending on $n$. – Dave Westwood Nov 23 '14 at 20:26
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Right. So, if $q_n$ is a primitive root modulo $2^n$, then $G = \langle q_n \rangle$, by definition of a primitive root. – Ben Grossmann Nov 23 '14 at 20:27
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You actually do get $G=\langle 5\rangle$. – Jyrki Lahtonen Nov 23 '14 at 20:43
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@Omnomnomnom Is the existence of primitive root $\mod 2^n$ is guaranteed? – Dave Westwood Nov 23 '14 at 20:58
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@DaveWestwood yes. See the link provided above. – Ben Grossmann Nov 23 '14 at 20:59
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Here lhf gives a link to a book containing a proof. As well as other bits. See also this answer. You should be able to prove that $5$ has the correct order using those ideas. – Jyrki Lahtonen Nov 23 '14 at 21:00