To determine which group $G = U(32)$ is isomorphic to by elimination, let's analyze the properties of $G$ and compare them with $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, and $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.

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Step 1: Determine the structure of $U(32)$

The group $U(32)$ is the group of units modulo $32$, meaning all integers $a$ such that $1 \leq a < 32$ and $\gcd(a, 32) = 1$. Let's compute:

• The factors of $32$ are $2^5$, so an element is coprime to $32$ if it is not divisible by $2$.
• The integers in $\{1, 2, \dots, 31\}$ that are coprime to $32$ are: $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31$
• There are $\phi(32) = 16$ such elements, so $|U(32)| = 16$.

Thus, $U(32)$ is a group of order $16$.

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Step 2: Check the structure of $H = \{1, 31\}$

The subgroup $H = \{1, 31\}$ is of order $2$ because:

• $31 \equiv -1 \pmod{32}$, so $31^2 \equiv 1 \pmod{32}$.
• Hence, $H \cong \mathbb{Z}_2$.

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Step 3: Compare with candidate groups

We now compare $U(32)$ with $\mathbb{Z}_8$, $\mathbb{Z}_4 \times \mathbb{Z}_2$, and $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.

Candidate 1: $\mathbb{Z}_8$

• $\mathbb{Z}_8$ is cyclic of order $8$. If $U(32) \cong \mathbb{Z}_8$, then $U(32)$ would have a generator of order $8$.
• Check whether $U(32)$ has elements of order $8$:
  • Using the structure of $U(32)$, it can be shown that $U(32)$ is not cyclic. Hence, $U(32) \not\cong \mathbb{Z}_8$.

Candidate 2: $\mathbb{Z}_4 \times \mathbb{Z}_2$

• $\mathbb{Z}_4 \times \mathbb{Z}_2$ has elements of orders $1, 2, 4$. If $U(32) \cong \mathbb{Z}_4 \times \mathbb{Z}_2$, there must be elements of orders $4$ and $2$.
• Check the orders of elements in $U(32)$. Indeed, $U(32)$ has elements of these orders.

Candidate 3: $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

• $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ has $2^4 = 16$ elements, all of order $1$ or $2$. If $U(32)$ is isomorphic to this group, then all elements of $U(32)$ must have order $1$ or $2$.
• Check the orders of elements in $U(32)$. There are elements of higher order (e.g., $3, 5$), so $U(32) \not\cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.

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Step 4: Conclude the isomorphism

From the above analysis:

• $U(32) \not\cong \mathbb{Z}_8$.
• $U(32) \not\cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$.
• $U(32) \cong \mathbb{Z}_4 \times \mathbb{Z}_2$.

Thus, $U(32) \cong \mathbb{Z}_4 \times \mathbb{Z}_2$.

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Would you like a breakdown of the orders of elements in $U(32)$, or additional clarification?

Here are some related questions:

1. How can we verify the structure of $U(32)$ computationally?
2. What is the significance of the orders of elements in identifying group isomorphism?
3. Why is $\mathbb{Z}_8$ not isomorphic to $U(32)$?
4. Can we generalize the method for finding isomorphisms for other $U(n)$?
5. How does the direct product structure $\mathbb{Z}_4 \times \mathbb{Z}_2$ manifest in group properties?

Tip: When solving group isomorphism problems, always analyze the orders of elements and the group's structure (e.g., cyclic or direct product).