We are tasked with solving a problem about the group $U(15)$, the group of units modulo 15 under multiplication. Let's tackle each part systematically:

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(i) What is the order of the group $U(15)$?

The group $U(15)$ consists of integers less than 15 that are coprime to 15 (i.e., integers $n$ such that $\gcd(n, 15) = 1$).

The Euler totient function, $\phi(15)$, gives the number of such integers:

$\phi(15) = \phi(3 \times 5) = \phi(3) \cdot \phi(5),$ where $\phi(3) = 2$ and $\phi(5) = 4$. Thus: $\phi(15) = 2 \cdot 4 = 8.$

So, the order of $U(15)$ is 8, and its elements are: $U(15) = \{1, 2, 4, 7, 8, 11, 13, 14\}.$

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(ii) Compute the order of 13 in $U(15)$.

The order of an element $a \in U(15)$ is the smallest positive integer $k$ such that: $a^k \equiv 1 \pmod{15}.$

For $a = 13$, we compute successive powers of 13 modulo 15:

1. $13^1 \equiv 13 \pmod{15}$,
2. $13^2 \equiv 169 \equiv 4 \pmod{15}$,
3. $13^3 \equiv 13 \cdot 4 \equiv 52 \equiv 7 \pmod{15}$,
4. $13^4 \equiv 13 \cdot 7 \equiv 91 \equiv 1 \pmod{15}$.

Thus, the order of 13 is 4.

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(iii) Find $\langle 13 \rangle$.

The subgroup $\langle 13 \rangle$ is the cyclic subgroup generated by 13. It consists of all powers of 13 modulo 15. From the calculations in part (ii): $\langle 13 \rangle = \{13^1, 13^2, 13^3, 13^4\} \pmod{15}.$ Substituting values: $\langle 13 \rangle = \{13, 4, 7, 1\}.$

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(iv) Show that $H = \{1, 11\}$ is a normal subgroup of $G$.

To check if $H$ is a normal subgroup, we need:

1. $H \subseteq U(15)$: Clearly, $1$ and $11$ are in $U(15)$.

2. Closure under multiplication modulo 15: $1 \cdot 1 = 1, \quad 1 \cdot 11 = 11, \quad 11 \cdot 11 = 121 \equiv 1 \pmod{15}.$ Hence, $H$ is closed under multiplication.

3. For normality: For any $g \in U(15)$ and $h \in H$, $g h g^{-1} \in H$.

   $g h g^{-1} \equiv g h g^{-1} \pmod{15}$ simplifies to an element in $H$, since $H$ is of order 2, and any subgroup of index 2 is normal.

Thus, $H = \{1, 11\}$ is a normal subgroup of $U(15)$.

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Summary of Results:

1. Order of $U(15)$: $8$.
2. Order of 13: $4$.
3. $\langle 13 \rangle$: $\{13, 4, 7, 1\}$.
4. $H = \{1, 11\}$ is a normal subgroup of $U(15)$.

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Do you need detailed explanations for any step?
Here are 5 related questions to expand your understanding:

1. What is the general method to compute $\phi(n)$ for composite numbers $n$?
2. Why does $g h g^{-1} \in H$ ensure normality of a subgroup?
3. How do cyclic subgroups relate to the structure of groups like $U(15)$?
4. Can we determine the structure of $U(15)$ (e.g., is it isomorphic to $\mathbb{Z}_n$)?
5. What are other examples of normal subgroups in small groups?

Tip: Understanding cyclic subgroups is crucial in modular arithmetic. Try constructing them explicitly to develop intuition!