4

Let $a$ be an odd integer and $n$ an integer such that $n\ge 3$.

1) I want to show that $a^{2^{n-2}} \equiv 1 \pmod{2^n}$

2) Then I want to show that $(\mathbb Z/{2^n\mathbb Z})^*$, the multiplicative group of integers modulo $2^n$, is not cyclic.

My Try:

1) Since $\phi(2^{n-1})=2^{n-2}$ and that $a$ is coprime with $2^{n-1}$, then by Euler's theorem we have that $a^{2^{n-2}} \equiv 1 \pmod{2^{n-1}}$ but I can't write the final conclusion.

2) I know that we have to show that no single element can generate the whole group but I don't know how to do it.

Thank you for your help!!

palio
  • 11,466

2 Answers2

3

Hints:

Suppose that you've a congruence like $f(x) \equiv a \pmod{p^m}$.

Do you know how you can construct a solution from this congruence for the congruence $f(x) \equiv a \pmod{p^{m+1}}$ ?

You have to use the Taylor formula for the expansion of $f(x+y)$ where $f$ is a polynomial. See what you get.

EDIT: Actually in this particular case, you can do it more easily by writing $a^{2^{k-2}} = 1 + t2^k$ and then applying the binomial theorem... But it's essentially the same thing and it's generally good to know how you can find the solutions of a congruence modulo a prime power provided that you've solved it for the previous exponent of that prime number.

For the second part, if $g$ is a generator of the group $(\mathbb{Z}/n\mathbb{Z})^\times$ where $n=2^k, k>2$, then the order of $g$ must be equal to the order of the group. So...

In elementary number theory, the fact that $(\mathbb{Z}/n\mathbb{Z})^\times$ is not cylic for $n=2^k, k>2$ is stated as there exist no primitive roots modulo $n$ for exponents of $2$ higher than three.

user66733
  • 7,449
  • Thank you for your help but How do you get $a^{2^{k-2}} = 1 + t2^k$ and what does this mean? – palio Nov 30 '13 at 23:21
  • @palio: We have $a^{2^{n-2}} \equiv 1 \pmod{2^n}$ by assumption. This is by definition the same as $2^n \mid a^{2^{n-2}} - 1$ which means that there exists an integer $t$ such that $a^{2^{n-2}} = 1 + t2^{n}$. I used $k$ instead of $n$. We are doing induction on $n$ and we wish to prove it for $n=k+1$. – user66733 Nov 30 '13 at 23:27
2

Hints:

For 1), start with $a^2 \equiv 1 \pmod{8}$ and induct on $n \geq 3$.

For 2), take a closer look at what you are proving in 1). What is the order of the group? You have already shown what you wanted to show: that no element generates the whole group.

Ivan Loh
  • 17,115