The image asks about the group of invertible elements in the ring $\mathbb{Z}_{32}$, along with other related questions about its structure.

Solution:

1. Invertible Elements in $\mathbb{Z}_{32}$: The invertible elements in $\mathbb{Z}_{n}$ are the integers $a$ such that $\gcd(a, n) = 1$. In this case, $n = 32$.

   The integers less than 32 and coprime to 32 are: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31$

   Thus, there are $\varphi(32) = 16$ invertible elements, where $\varphi$ is the Euler totient function.

2. Structure of $U(\mathbb{Z}_{32})$: The group of invertible elements, $U(\mathbb{Z}_{32})$, is isomorphic to: $U(\mathbb{Z}_{32}) \cong \mathbb{Z}_8 \times \mathbb{Z}_2$ This decomposition follows from the Chinese Remainder Theorem and the structure theorem for finite abelian groups.

3. Non-invertible Elements in $\mathbb{Z}_{32}$: The non-invertible elements are the elements $a$ such that $\gcd(a, 32) > 1$. These include: $0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30$

4. Do Non-invertible Elements Form an Ideal? Yes, the non-invertible elements form the ideal generated by 2 in $\mathbb{Z}_{32}$, denoted as $(2)$. This is because $(2) = \{ 2k \mod 32 \mid k \in \mathbb{Z} \}$, which includes all non-invertible elements.

Final Answers:

1. Group of invertible elements: $U(\mathbb{Z}_{32}) \cong \mathbb{Z}_8 \times \mathbb{Z}_2$ (Abelian, not cyclic).
2. Non-invertible elements: $0, 2, 4, \dots, 30$.
3. Non-invertible elements form an ideal? Yes.

Do you want additional clarification on these concepts?

Here are 5 related questions for deeper understanding:

1. Why does the Chinese Remainder Theorem lead to the decomposition of $U(\mathbb{Z}_{32})$?
2. How do we compute the Euler totient function $\varphi(n)$ in general?
3. What are the differences between cyclic and non-cyclic abelian groups?
4. How does the structure theorem for finitely generated abelian groups apply here?
5. What are other examples of rings where invertible elements form interesting groups?

Tip: When finding invertible elements in $\mathbb{Z}_n$, always check $\gcd(a, n) = 1$ to ensure $a$ has an inverse modulo $n$.