Tips for classifying $ (\mathbb{Z} / 2^n \mathbb{Z})^\times$ To finish off a problem earlier, I had to classify the group $G =(\mathbb{Z} / 16 \mathbb{Z})^\times$.
I said $G$ is abelian with order $\phi(16) = 8$, and found three order-2 elements (namely $7^2 = 9^2 = 15^2 = 1$), thus it's $C_2 \times C_2 \times C_2$ (using the Structure Theorem, since it can't be $C_2 \times C_4$ or $C_8$). EDIT: Actually, this means it's $ C_2 \times C_4 $, my bad.
I could only really do this since we have a small modulus like $2^4$, but I'm interested as to whether there a general approach for $2^n$? Of course for odd primes $p \ge 3$, we can use the theory of primitive roots, but this doesn't apply for powers of 2. It seems like this is a fairly important problem, since it arises quite naturally when decomposing $(\mathbb{Z}/m\mathbb{Z})^\times$ for general $m$.
 A: Note that $${\displaystyle \{\pm 1,2^{n-1}\pm 1\}\cong \mathrm {C} _{2}\times \mathrm {C} _{2},}\{\pm 1, 2^{n-1} \pm 1\}\cong \mathrm{C}_2 \times \mathrm{C}_2,$$ is the
$2-$torsion subgroup (so ${\displaystyle (\mathbb {Z} /2^{n}\mathbb {Z} )^{\times }}$  is not cyclic) and the powers of 3 are a cyclic subgroup of order $2^{n − 2},$ so we have
$${\displaystyle (\mathbb {Z} /2^{n}\mathbb {Z} )^{\times }\cong \mathrm{C}_2 \times \mathrm{C}_{2^{n-2}}}.$$
See https://en.wikipedia.org/wiki/Torsion_subgroup as well as https://en.wikipedia.org/wiki/Multiplicative_group_of_integers_modulo_n
A: For $n ≥ 2$, ${\displaystyle (\mathbb {Z} /2^{n}\mathbb {Z} )^{\times }}$  is the internal direct product of the subgroup of order 2 generated by the class of $–1$ and
the subgroup of order $2^{n-2}$ generated by the class of $5.$
See a proof (in french) in Anneau $ℤ/nℤ$ → Cas où $n$ n'est pas premier → Si $p = 2$.
In particular for $n=4$, it is not (isomorphic to) $C_2\times C_2\times C_2$ as you claim, but to $C_2\times C_4.$ The three elements of order $2$ you found are (modulo $16$): $-1\equiv15$, $5^2\equiv9,$ and $-(5^2)\equiv-9\equiv7.$
