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Denote by $(\mathbb{Z}/2^n\mathbb{Z})^\times$ the multiplicative group of odd congruence classes modulo $2^n$. I need to find for which $n$ this multiplicative group is cyclic. This is homework, so I would appreciate a general direction or a simple outline instead of a full answer. Here's my work so far:

By computing $(\mathbb{Z}/2^n\mathbb{Z})^\times$ for $n$ up to 4, I claimed that the group is not cyclic for $n \geq 3$, which I confirmed later on Wikipedia. I also received a (not proven) formula from the exercise: if $a$ is odd, then $a^{2^{n-2}} \equiv 1 \pmod{2^n}$ for $n \geq 3$. I understood that this formula is a upper bound on the order of the elements of $(\mathbb{Z}/2^n\mathbb{Z})^\times$: it must divide $2^{n-2}$, so it cannot be $\varphi(2^n) = 2^{n-1}$. This basically proves the exercise, but I don't feel like I should use this statement without a proof - it would make this exercise too direct comparing with any other from our exercise list.

So then I went out to prove the formula. I started with Euler's Theorem: $$a^{2^{n-1}} \equiv 1 \pmod{2^n}$$ Then I took the square root: $$a^{2^{n-2}} \equiv 1 \pmod{2^n}\\ \text{or } a^{2^{n-2}} \equiv -1 \pmod{2^n}$$ So I tried to prove the second equation false, but I couldn't get far by doing simple arithmetic and factorization, like $a^{2^{n-2}} \equiv 1 + 2 + 4 + ... + 2^{n-1} \pmod{2^n}$. Induction over $n$ doesn't seem to be the case. What am I missing?

EDIT: I just fiddled around more and now I believe induction is the way. The induction hypotesis would be $a^{2^{n-3}} \equiv 1 \pmod{2^{n-1}}$, which means $a^{2^{n-3}} = 2^{n-1}k + 1$ for some integer $k$. By squaring, I have $a^{2^{n-2}} = 2^{2n - 2}k^2 + 2^nk + 1 \equiv 1 \pmod{2^n}$.

Henrique Augusto Souza
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2 Answers2

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You can prove that for any $a\in (\mathbb Z/2^n\mathbb Z)^\times$, $a^{2^{n-2}}\equiv 1 \pmod {2^n}$ by induction.

The inductive step works as follows. Suppose that for any $a\in (\mathbb Z/2^n\mathbb Z)^\times$, $a^{2^{n-2}}\equiv 1 \pmod {2^n}$, and let $b\in (\mathbb Z/2^{n+1}\mathbb Z)^\times$. Then by induction $$b^{2^{n-2}}\equiv1\pmod {2^n}$$ and hence $$b^{2^{n-2}}\equiv 1 +c2^n\pmod {2^{n+1}}$$for some $c$. Now observe that $b^{2^{n-1}} = (b^{2^{n-2}})^2$. Can you finish from here?

Mathmo123
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$\textbf{Hint}$: You can try to prove that for $n \geq 3$, $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2^{n-2}\mathbb{Z} \cong (\mathbb{Z}/2^{n}\mathbb{Z})^{\times}$, where the isomorphism is given by mapping $(\overline{i}, \overline{j})$ to $\overline{(-1)}^{i} \overline{5}^j$. You have to check that this is a well-defined homomorphism. For proving the bijetivity note that it will be enough to prove injectivity for reasons of equal (finite) cardinality.

Plankton
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  • A more algebraic approach than I was thinking, but our course in Number Theory 1 also uses alot of deep algebra results to illustrate how powerful abstraction lies underneath elementary theorems. For instance, we went all the way to Lagrange's theorem to prove Fermat's Little Theorem for the first time, or to Gaussian Integers to prove Fermat's Christmas Theorem. I will give a try on this approach too later :) – Henrique Augusto Souza Nov 18 '15 at 18:26