I need to show that the group $U_{16}$ is isomorphic to the group $\mathbf{Z_4} \times \mathbf{Z_2}$.
I could clearly write out the multiplication tables for the two groups and then find a mapping of each element from $U_{16}$ to each element of $\mathbf{Z_4} \times \mathbf{Z_2}$, but this seems like too many computations. Since each group is of order $8$, their multiplication tables would each consist of $64$ computations, and seems like a boring way to do this problem.
Is there a smarter/more elegant way of doing so?
Also, since I am told that they are isomorphic, the two groups must share invariant properties, i.e., being abelian, being cyclic etc.
Thank you for your help and time!
You're right that there are slicker ways to do this. Here's one sort of follow-your-nose approach:
You know that $U(16)$ consists of all (equivalence classes mod $16$ of) the odd integers $\{1,3,5,\ldots,15\}$, which is a set of order $8$. You can start by thinking about how different powers of a prime behave in this group. For instance, you can check $\langle 3 \rangle$ is a subgroup of order $4$, as is $\langle 5 \rangle$, and $\langle 7 \rangle$ is a subgroup of order $2$ that intersects the previous two subgroups trivially. This implies (why?) that, for instance, $\{3^m 7^n : 0\leq m \leq 3, 0 \leq n \leq 1\}$ is of order $8$, and that $(m,n) \mapsto 3^m7^n$ defines a surjective homomorphism $\mathbb Z_4 \times \mathbb Z_2 \to U(16)$.
Edit: Whoops! I missed the conversation in the comments. You've found an analogous argument using instead $(m,n) \mapsto 5^m 7^n$ instead.