$U(2^n) \thickapprox Z_2\oplus Z_{2^{n-2}}$ Show that $$U(2^n) \thickapprox Z_2\oplus Z_{2^{n-2}}\,\,\,\,\,\,,\text{for }n\geq3$$

Well I think this can be done with the help of this theorem
Let $m=n_1n_2.\,.\,.\,.n_k$ where $\gcd(n_i,n_j)=1$ for $i\neq j.$ Then,
$$U(m) \thickapprox U(n_1)\,\oplus U(n_2)\, \oplus .\,.\,.\,.\oplus\, U(n_k) $$
 A: The theorem you are quoting does not help. Instead, show that 


*

*$-1$ has order $2$ in $U(2^{n})$,

*$5$ has order $2^{n-2}$, 

*the subgroup generated by $5$ does not contain $-1$ .

A: Put $\;k=2^n\;$ for simplicity, and observe that
$$\begin{align*}&\left(2^k-1\right)^2=2^{2k}-2^{k+1}+1=1\pmod{2^k}\\{}\\&\left(2^{k-1}-1\right)^2=2^{2k-2}-2^k+1=1\pmod{2^k}\end{align*}$$
The last equivalence because $\;2k-2\ge k\iff k\ge 2\;$, which is the case we're interested in.
Thus, since $\;2^{k-1}-1\neq 2^k-1=-1\pmod{2^k}\;$, there are already two different elements of order two in $\;U(2^k)\;$ , and this means this group can't be cyclic.
Now you can prove  by induction on $\;n\;$ that
$$n\ge3\implies 5^{2^{n-2}}=1+r2^n\;,\;\;r\;\;\text{odd}$$
when for the Inductive Step you can go as
$$5^{2^{(n+1)-2}}=5^{2^{n-1}}=5^{2^{n-2}}\cdot5^{2^{n-2}}\stackrel{\text{I.S.}}=(1+r2^n)^2=1+(r+r^22^{n-1})2^{n+1}...\text{etc.}$$
Also observe that it can't be $\;5^{2^{n-3}}=1\pmod {2^n}\;$ otherwise by what we already proved above we'd get that for odd $\;s\;$ :
$$1+r2^n=\overbrace{1+s2^{n-1}}^{=5^{2^{n-3}}\;\text{by the above}}\implies2r=s\;,\;\;\;\text{contradiction}$$
and from here that $\;\text{ord}_{2^n}\,5=n-2\;$
Try now to complete the proof using the Fundamental Theorem of Finite Abelian Groups.
