$(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic Group for $n\geq 3$

Question

Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic Group for $n\geq 3$.

Hint : Find two subgroups of order $2$.

I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof.

I have no idea how to proceed for this.

any hint would be appreciated.

Thank You.

Answer 1

Follow this outline:

• $5$ has order $m=2^{n-2}$. For a proof, see https://math.stackexchange.com/a/74086/589 .

• $5^r$ and $-5^r$ for $r=2^{n-3}$ generate different subgroups of order $2$.

If you only want to prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not cyclic, it is enough to prove that no element can have order $2^{n-1}$. For a proof, see How to prove by induction that $a^{2^{k-2}} \equiv 1\pmod {2^k}$ for odd $a$?.

Answer 2

Making lhf's fine (+1) answer perhaps a bit more concrete. There are three subgroups of order two: $H_1=\{1,-1\}$, $H_2=\{1,2^{n-1}+1\}$ and $H_3=\{1,2^{n-1}-1\}$. The non-1 element in each subgroup has square $\equiv1\pmod{2^n}$ as expected.

Answer 3

Here is a simple way to do this :

1. $U(8) = \{[1],[3],[5],[7]\}$, and check that $$ [3]^2 = [5]^2 = [7]^2 = [1] $$ so $U(8)$ is not cyclic (it doesn't have an element of order $4 = |U(8)|$)

2. Since $8 \mid 2^n$ for $n > 3$, we have a natural ring homomorphism $$ \mathbb{Z}/2^n\mathbb{Z} \to \mathbb{Z}/8\mathbb{Z}, \text{ given by } [x]_{2^n} \mapsto [x]_8 $$ which induces a surjective group homomorphism $$ U(2^n) \to U(8) $$ Since the quotient of a cyclic group must be cyclic, it follows that $U(2^n)$ cannot be cyclic for $n\geq 3$

Answer 4

Consider $\pm(2^{k-1}+1)$. What is the order of these two elements if $k\geqslant 3$? Note they are $\neq \pm 1$ if $k\geqslant 3$.