Theorem 3

If $a = \prod p_i^{n(i)}$ is the prime decomposition of the integer $a > 1$, then the group $(\mathbf{Z}/a\mathbf{Z})^*$ of invertible elements of the ring $\mathbf{Z}/a\mathbf{Z}$ is isomorphic to the product of the groups $(\mathbf{Z}/p_i^{n(i)}\mathbf{Z})^*$. If $p > 2$ is a prime number and $n \geq 1$ an integer, then the group $(\mathbf{Z}/p^n\mathbf{Z})^*$ is cyclic of order $p^n \ ^1(p - 1)$. The group $(\mathbf{Z}/2\mathbf{Z})^*$ is trivial; for $n \geq 2$ the group $(\mathbf{Z}/2^n\mathbf{Z})^*$ is the direct product of the cyclic group of order $2^n \ ^2$ generated by the residue class of 5 mod $2^n$ and the cyclic group of order 2 consisting of the residue classes of 1 and $-1$ mod $2^n$.

The orders $p^n \ ^1$ of $U(p^n)$ and $p - 1$ of $(\mathbf{Z}/p\mathbf{Z})^*$ are coprime; since $U(p^n)$ and $(\mathbf{Z}/p\mathbf{Z})^*$ are cyclic (Prop. 3 and V, p. 78, Lemma 1), the group $(\mathbf{Z}/p^n\mathbf{Z})^*$ is cyclic (apply Lemma 4 to the exact sequence (3)). If $n \geq 2$ then the restriction of the homomorphism $v : (\mathbf{Z}/2^n\mathbf{Z})^* \to (\mathbf{Z}/4\mathbf{Z})^*$ to the subgroup $(1, -1)$ is bijective; the group $(\mathbf{Z}/2^n\mathbf{Z})^*$ is thus the direct product of this subgroup and the kernel $U(2^n)$ of $v$; the result follows from Prop. 3.