$(\mathbb{Z}/2^n \mathbb{Z})^*$ 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.
 A: 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.
A: Here is a simple way to do this :


*

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

*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 $n\geq 3$
A: 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 $a$?.
A: 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 $k\geqslant 3$. 
