5

Let G be the multiplicative group $\mathbb{Z}_n^*$ for $n = 2^k$ and $k \ge 3$.

Can we prove that no element has order bigger than $2^{k-2}$ ?

My solution (not really a solution) :

Since $n=2^k$, I thought that this group is cyclic and the generator is 2. I'm not sure if the group order can't be bigger than $2^{k-2}$?

zyx
  • 35,436
baaa12
  • 754
  • Perhaps I misunderstand the notation, but I would have thought $\Bbb{Z}_n^*$ referred to the multiplicative group of elements in the ring $\Bbb{Z}_n$ which are relatively prime to $n$, so that for $n$ a power of two, it wouldn't be true that "the generator is 2". – hardmath Apr 04 '13 at 15:10
  • The group of multiplicative units of $\mathbb{Z}/8\mathbb{Z}$ is the Klein four group, which is a counterexample to the claim on order and that the group is cyclic. – Jared Apr 04 '13 at 15:11
  • Yes I probably wrong and it's not even generator... – baaa12 Apr 04 '13 at 15:11
  • In fact, we can use Euler's totient function to find the exact order of the group: $\varphi(2^k)=2^{k-1}$. In this case, all this is saying is that the multiplicative units of $\mathbb{Z}/n\mathbb{Z}$ are exactly the odd elements, which is why we divide by a factor of 2. – Jared Apr 04 '13 at 15:12
  • You may be thinking of the field of order $2^k$, the units of which do form a cyclic multiplicative group, but these are not the same thing. – Cameron Buie Apr 04 '13 at 15:13
  • @Jared Thanks I tried this but I dont understand why is cant be bigger than 2^k-2 this way? – baaa12 Apr 04 '13 at 15:16
  • $\forall a \in \Bbb Z / 2^k\Bbb Z,a\in G \Leftrightarrow a \land 2^k = 1 \Leftrightarrow a \land 2 = 1 \Leftrightarrow 2 \not \mid a$ so you remove exactly half of the elements and hence get a group of oder $2^{k-1}$ – xavierm02 Apr 04 '13 at 15:18
  • Perhaps the issue here is that the group G is not cyclic, and that therefore the bound on order of any element is strictly smaller than the order of the group itself. – hardmath Apr 04 '13 at 15:23
  • See also here – Bill Dubuque May 28 '23 at 19:25

3 Answers3

3

An element $a\in\mathbb{Z}/n\mathbb{Z}$ is a unit if and only if $(a,n)=1$, that is, if and only if $a$ and $n$ are relatively prime. In your case, since the only prime divisor of $n=2^k$ is $2$, this equivalence reduces to $a$ is a unit if and only if $a$ is odd. Hence, the elements of $(\mathbb{Z}/2^k\mathbb{Z})^*$ are $$1,3,5,7,\ldots,2^k-5,2^k-3,2^k-1$$ of which there are $2^{k-1}$, or just half of $2^k$ because we are only taking odd elements. Now $(\mathbb{Z}/2^k\mathbb{Z})^*$ is not cyclic for $k\ge 3$ (read here, which states that $(\mathbb{Z}/2^k\mathbb{Z})^*\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2^{k-2}\mathbb{Z}$). It follows that any element $a$ must have order less than $2^{k-1}$, but still dividing $2^{k-1}$, and hence must be no larger than $2^{k-2}$.

Jared
  • 31,451
3

The lowbrow explanation is that squares of odd numbers are equal to $1$ mod $8$. This is 1/4, not 1/2, of the elements in the multiplicative group, so that squaring (which is doubling, in the group) does twice as much collapsing of the residue classes than if that group were cyclic. This collapsing persists for higher powers of $2$ as one can always reduce any such situation mod $8$.

This is related to the fact that there are complications in "lifting" of solutions to $x^2=a$, from $\mod 2^k$ to $\mod 2^{k+1}$, but not for lifting the exponent in a solution mod $p^u$ for odd $p$.

zyx
  • 35,436
2

If we simply want to show that the multiplicative group $G = \Bbb{Z}_{2^k}^*$ is not cyclic for $k \ge 3$ (which is all that is needed to conclude the order of any element is strictly less than the order of group $G$), then we can verify it first for the case $k = 3$ as Jared's original comment points out: $\Bbb{Z}_8^*$ is isomorphic to the product of two cyclic groups of order two (and has no element of order 4).

Then for $k \gt 3$ one verifies that the natural ring homomorphism from $\Bbb{Z}_{2^k}$ onto $\Bbb{Z}_8$ (taking the coarser congruence relation) induces a group homomorphism of the respective multiplicative groups (the groups of units of the respective rings). From this we see that $\Bbb{Z}_{2^k}^*$ cannot be cyclic, since the quotient group of a cyclic group is again cyclic.

The specific structure of $\Bbb{Z}_{2^k}^*$ can be worked out from the fact that it has two cyclic factors, $C_2 \times C_{2^{k-2}}$, with a generator for the second factor being 3 in all orders ($k \ge 3$). A write-up in terms of "primitive roots" is given here in Lemma 12, where it is shown by induction that for any odd $a$:

$$ k \ge 3 \implies a^{2^{k-2}} \equiv 1 \mod 2^k $$

hardmath
  • 37,015