Let's work through each part of the problem step by step.

1. The order of the group $U(15)$

The group $U(15)$ refers to the group of units modulo 15. These are the integers less than 15 that are relatively prime to 15 (i.e., the integers $x$ such that $\gcd(x, 15) = 1$).

To determine the elements of $U(15)$, we find the integers less than 15 and check their gcd with 15:

• $\gcd(1, 15) = 1$
• $\gcd(2, 15) = 1$
• $\gcd(3, 15) = 3$ (not coprime)
• $\gcd(4, 15) = 1$
• $\gcd(5, 15) = 5$ (not coprime)
• $\gcd(6, 15) = 3$ (not coprime)
• $\gcd(7, 15) = 1$
• $\gcd(8, 15) = 1$
• $\gcd(9, 15) = 3$ (not coprime)
• $\gcd(10, 15) = 5$ (not coprime)
• $\gcd(11, 15) = 1$
• $\gcd(12, 15) = 3$ (not coprime)
• $\gcd(13, 15) = 1$
• $\gcd(14, 15) = 1$

Thus, the elements of $U(15)$ are $1, 2, 4, 7, 8, 11, 13, 14$.

The order of the group $U(15)$ is the number of elements in the group. Hence, the order of the group is: $\text{Order of } U(15) = 8$

2. The order of 13 in $U(15)$

The order of an element $a$ in a group is the smallest integer $k$ such that $a^k \equiv 1 \ (\text{mod} \ 15)$. Let's compute the powers of 13 modulo 15 until we get 1.

• $13^1 \equiv 13 \ (\text{mod} \ 15)$
• $13^2 \equiv 13 \times 13 = 169 \equiv 169 \mod 15 = 169 - 11 \times 15 = 169 - 165 = 4$
• $13^3 \equiv 13 \times 4 = 52 \equiv 52 \mod 15 = 52 - 3 \times 15 = 52 - 45 = 7$
• $13^4 \equiv 13 \times 7 = 91 \equiv 91 \mod 15 = 91 - 6 \times 15 = 91 - 90 = 1$

Thus, the order of 13 in $U(15)$ is 4 because $13^4 \equiv 1 \ (\text{mod} \ 15)$.

3. The subgroup $\langle 13 \rangle$

The subgroup generated by 13 is the set of all powers of 13 modulo 15. From the previous calculations, we have:

• $13^1 \equiv 13 \ (\text{mod} \ 15)$
• $13^2 \equiv 4 \ (\text{mod} \ 15)$
• $13^3 \equiv 7 \ (\text{mod} \ 15)$
• $13^4 \equiv 1 \ (\text{mod} \ 15)$

Therefore, the subgroup $\langle 13 \rangle$ is: $\langle 13 \rangle = \{ 1, 4, 7, 13 \}$

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

To show that $H = \{1, 11\}$ is a normal subgroup of $U(15)$, we need to check two things:

1. Subgroup condition: $H$ must be a subgroup of $U(15)$.

   • $1 \in U(15)$ and $11 \in U(15)$, so $H = \{1, 11\}$ is a subset of $U(15)$.
   • $H$ is closed under multiplication modulo 15:
     • $1 \times 1 = 1$, $1 \times 11 = 11$, $11 \times 11 = 121 \equiv 1 \ (\text{mod} \ 15)$.
     • So, $H$ is closed under multiplication.
   • $H$ contains the identity element (1) and is closed under inverses, so it is indeed a subgroup.

2. Normality: A subgroup $H$ of $G$ is normal if for every element $g \in G$, the conjugate of any element of $H$ by $g$ is still in $H$, i.e., for all $h \in H$ and $g \in G$, $g h g^{-1} \in H$.

Let's check the conjugates of the elements of $H$ by all elements of $U(15)$:

• For $g = 1$, $1 \times 1 \times 1^{-1} = 1$, and $1 \times 11 \times 1^{-1} = 11$.
• For $g = 2$, $2 \times 1 \times 2^{-1} = 2 \times 1 \times 8 = 8 \neq 1, 2 \times 11 \times 2^{-1} = 2 \times 11 \times 8 \equiv 11$.
• $H = \{ 1, 11\}$.

Therefore,