I have to determine the number of elements of order $≤ 2$ in $(\mathbb{Z}/2^n\mathbb{Z})^×$, and use this to find the rank and the elementary divisors of $(\mathbb{Z}/2^n\mathbb{Z})^×$
I know that $(\mathbb{Z}/2^n\mathbb{Z})^× \rightarrow (\mathbb{Z}/4\mathbb{Z})^×$ is a well defined surjective homomorphism
Any help would be grateful.
Hint: Prove by induction that $5^{2^{n−2}} \equiv 1 \bmod 2^n$ and $5^{2^{n−3}} \equiv 1+2^{n-1} \bmod 2^n$ for $n\ge 3$.
We know that: $$ \begin{cases} 5^{2^{n-2}}\equiv1\pmod{2^n}\\ \begin{aligned} \big|(\mathbb{Z}/2^n\mathbb{Z})^{\times}\big|&=\varphi(2^n)\\ &=2^{n-1}(2-1)\\ &=2^{n-1} \end{aligned}\\ 5^{2^{n-3}}\equiv2^{n-1}+1\pmod{2^n}\\ \begin{aligned} -5^{2^{n-3}}&\equiv2^n-(2^{n-1}+1)\pmod{2^n}\\ &=2^{n-1}-1\pmod{2^n} \end{aligned}\\ 2^n-1\equiv-1\pmod{2^n}\\ \end{cases} $$ Hence: $$ (\mathbb{Z}/2^n\mathbb{Z})^{\times}\cong\langle5\rangle\times\langle-1\rangle=\mathbb Z_{2^{n-2}}\times\mathbb Z_2. $$ A group $G$ is cyclic if and only if it possesses a unique subgroup of each order dividing $|G|$ .
Since both $\langle5\rangle$ and $\langle-1\rangle$ are cyclic, $(2^{n-1}+1)$ and $(-1)$ is the unique element of order $2$ in $\mathbb Z_{2^{n-2}}$ and $\mathbb Z_2$, respectively.
Thus, the elements of order smaller than $2$ in $(\mathbb{Z}/2^n\mathbb{Z})^{\times}$ and $\mathbb Z_{2^{n-2}}\times\mathbb Z_2$ are: $$ \left.\begin{cases} \begin{gathered} 2^{n-1}+1\ ,\\-1\ ,\\2^{n-1}-1\ ,\\1 \end{gathered} \end{cases}\right\}\cong \left.\begin{cases} \begin{gathered} (2^{n-1}+1\,,1)\ ,\\(1\,,-1)\ ,\\(2^{n-1}+1\,,-1)\ ,\\(1\,,1) \end{gathered} \end{cases}\right\}\ , $$ respectively, and they form a group which is isomorphic to $\mathbf{V}_4$ the Klein $4$-group.
