Relation Between $(\mathbb{Z}/a\mathbb{Z})^\times$ and $(\mathbb{Z}/ab\mathbb{Z})^\times$ How can we prove that if there exists an element of order $c$ in $(\mathbb{Z}/a\mathbb{Z})^\times$ then there must exist some element of order $c$ in $(\mathbb{Z}/ab\mathbb{Z})^\times$?
 A: This follows easily from the fact that $(\mathbb Z/p^m\mathbb Z)^\times$ is (isomorphic to) a subgroup of $(\mathbb Z/p^n\mathbb Z)^\times$ whenever $m \le n$.  A brief sketch of this claim follows.
For $p$ odd, both groups are cyclic and the order of one divides the other.  For $p=2$, the group $(\mathbb Z/2^m\mathbb Z)^\times$ is isomorphic to $\mathbb Z/2 \times \mathbb Z/2^{m-2}\mathbb Z$ provided $m \ge 2$, so again the claim holds for $2 \le m \le n$.  When $p=2$ and $m=1$ the claim is trivial.
A: It seems like you're comfortable with the reductions needed to apply Erick's answer. A sketch for the curious reader: (1) Chinese remainder theorem (2) $(A \times B)^* = A^* \times B^*$ (3) If $x$ and $y$ are elements of a group which commute, then the order of $xy$ is the l.c.m. of the orders of $x$ and $y$.
Here's a slightly different argument for the last step. If $n \geq m$ are natural numbers then the reduction map $\mathbb Z/p^n\mathbb Z \to \mathbb Z/p^m\mathbb Z$ induces a surjective map on unit groups $(\mathbb Z/p^n\mathbb Z)^* \to (\mathbb Z/p^m\mathbb Z)^*$. [Why?] If you have an element of order $k$ in $(\mathbb Z/p^m\mathbb Z)^*$ then its preimages under this map have orders divisible by $k$.
