Let's break down the problem step by step. We are given a group $G = H \times H$, where $H = \mathbb{Z}/15\mathbb{Z}$ under addition. We need to analyze the function $f: G \rightarrow H$ defined by $f(a, b) = a + b$, where $a, b \in H$.

Part 1: Proving that $f$ is a Group Homomorphism

A map $f: G \rightarrow H$ is a group homomorphism if for all $(a_1, b_1), (a_2, b_2) \in G$, the following holds: $f((a_1, b_1) + (a_2, b_2)) = f(a_1, b_1) + f(a_2, b_2).$

First, note that in $G$, the addition operation is defined as: $(a_1, b_1) + (a_2, b_2) = (a_1 + a_2, b_1 + b_2),$ where $+$ is addition modulo 15.

Now, let’s compute each side of the homomorphism condition:

1. Left-hand side: $f((a_1, b_1) + (a_2, b_2)) = f(a_1 + a_2, b_1 + b_2).$ By definition of $f$: $f(a_1 + a_2, b_1 + b_2) = (a_1 + a_2) + (b_1 + b_2).$ In $H = \mathbb{Z}/15\mathbb{Z}$, this sum is taken modulo 15.

2. Right-hand side: $f(a_1, b_1) + f(a_2, b_2) = (a_1 + b_1) + (a_2 + b_2).$ Again, the sum is taken modulo 15.

Notice that both sides are equal because addition in $\mathbb{Z}/15\mathbb{Z}$ is associative and commutative. Therefore, $f$ is indeed a homomorphism.

Part 2: Determining $N = \ker(f)$

The kernel of $f$ is defined as: $N = \ker(f) = \{ (a, b) \in G \mid f(a, b) = 0 \}.$ This means we need to find all pairs $(a, b) \in \mathbb{Z}/15\mathbb{Z} \times \mathbb{Z}/15\mathbb{Z}$ such that $a + b \equiv 0 \pmod{15}$.

In other words, $a \equiv -b \pmod{15}$. The elements of $N$ are: $N = \{ (a, -a) \mid a \in \mathbb{Z}/15\mathbb{Z} \}.$

There are 15 such elements since $a$ can take any value in $\mathbb{Z}/15\mathbb{Z}$. So, $N = \{ (0, 0), (1, 14), (2, 13), \dots, (14, 1) \}.$

Part 3: Subgroups of the Quotient Group $G/N$

The quotient group $G/N$ consists of cosets of $N$ in $G$. Each coset is of the form: $(a, b) + N \text{ for } (a, b) \in G.$

To understand the structure of $G/N$, let’s determine the order of the quotient group. The order of $G$ is $15 \times 15 = 225$. The order of $N$ is 15, so the order of $G/N$ is: $\frac{225}{15} = 15.$

The quotient group $G/N$ is isomorphic to $\mathbb{Z}/15\mathbb{Z}$. This is because $G/N$ has 15 elements, and the operation is addition modulo 15.

Conclusion

• $f$ is a group homomorphism.
• $\ker(f)$ is $\{ (a, -a) \mid a \in \mathbb{Z}/15\mathbb{Z} \}$.
• The quotient group $G/N$ is isomorphic to $\mathbb{Z}/15\mathbb{Z}$.

Follow-Up Questions

1. What are the possible orders of subgroups of $G/N$?
2. Can you describe the cyclic subgroup structure within $G/N$?
3. How would the structure change if $H = \mathbb{Z}/n\mathbb{Z}$ for some $n \neq 15$?
4. What is the center of the group $G = H \times H$?
5. How would you generalize this analysis for $H \times H \times H$?

Tip

When determining the quotient group $G/N$, always start by calculating the orders of $G$ and $N$ to understand the size and structure of the resulting quotient group.