Isomorphism of Quotient Group: $U(32)/U_{16}(32)$ Let $U(32)$ be the group of numbers relatively prime to $32$ under multiplication modulo $32$. The set $U(32)$ will then consist of the elements $(1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23,
25, 27, 29, 31)$. Let $U_{16}(32)$ be the subgroup of $U(32)$ that are equal to $1$ modulo $16$. $U_{16}(32)$ will then consist of the elements $(1,17)$.
The book I am reading asserts that all finite Abelian groups are products of cyclic groups (to be proven later). They then show how $U(32)/U_{16}(32) = \mathbf{Z}_4 \oplus \mathbf{Z}_2$ and not $\mathbf{Z}_2 \oplus \mathbf{Z}_2 \oplus \mathbf{Z}_2$ or $\mathbf{Z}_8$ because $U(32)/U_{16}(32)$ has an element of order $4$. I'm skipping some steps -- but I was able to follow until here.
Now they state:

This proves that $U(32)/U_{16}(32) = \mathbf{Z}_4 \oplus \mathbf{Z}_2$, which (not so incidentally!) is isomorphic to $U(16)$.

How do I see this last statement?
 A: To understand why $U(32)/U_{16}(32) \cong U(16)$, note the following remarks. An integer is coprime to $32$ if and only if it's comprime to $16$. So in fact you can consider the $U(32)$ as the set of integers coprime to $16$ and less than $32$. 
Now note that if $a \in U(32)$ then we have that $16+a \in U(32)$. So as we want to reduce $U(32)$ to the set of integers comprime to $16$ and less than $16$ we just need to take everything modulo $16$, which is exactly what $U_{16}(32)$ does. If you want to be more rigorous you can invoke the Isomorphism Theorems. From what we said in the first paragraph we have that a natural homomorphism $\phi: U(32) \to U(16)$ given by $\phi(x_{32}) = x_{16}$. Now $\ker \phi = \{x \in U(32) | x \equiv 1 \pmod {16}\} = U_{16}(32)$. So by the first isomorphism theorem:
$$U(16) =\phi[U(32)] \equiv U(32)/\ker \phi = U(32)/U_{16}(32)$$
Now it's not hard to conclude that $U(16) = \mathbf{Z}_4 \oplus \mathbf{Z}_2$, as it has no elements of order $8$, but it has an element of order $4$.
