TITLE: Why multiplicative group $\mathbb{Z}_n^*$ is not cyclic for $n = 2^k$ and $k \ge 3$
QUESTION [5 upvotes]: 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 $2^{k-2}$?

REPLY [2 votes]: 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 $p$.