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

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.

Solution 1:

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.

Solution 2:

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 $n\geq 3$