How to prove that $U_{2^n}$ is isomorphic as group to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$ for $n \ge 3$?

Question

How to prove that $U_{2^n}$ is isomorphic as group to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$ for $n \ge 3$ ?

Answer 1

You prove that $5$ has order $2^{n-2}$ by induction.

Answer 2

Hint : Show that for $n \geq 3$, $5$ has order $2^{n-2}$, arguing as follows: by induction, you can prove that $5^{2^{n-3}} \equiv 1 + 2^{n-1}$ mod $2^{n}$. Now you know that $-1 \equiv 2^{n}-1$ has order 2. Using this, you can prove that every element can be written uniquely as $(-1)^{a}5^{b}$, where $a \in \lbrace 1,0\rbrace$ and $0 \leq b <2^{n-2}$. Then you construct an isomorphism, sending $(-1)^{a}5^{b} \to (a,b) \in \mathbf{Z}/2\mathbf{Z} \times \mathbf{Z}/2^{n-2}\mathbf{Z}$