Hw10 Problem 2

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### Problem 2

(+) Prove that the mapping $\phi : \mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$ given by $(a, b) \mapsto a - b$ is a homomorphism. What is the kernal of $\phi$? Describe the set $\phi^{-1}(3)$ (that is, all elements that map to $3$).

Solution

Proof: Let $u, v \in \mathbb{Z} \times \mathbb{Z}$ where $u = (a, b)$ and $v = (c, d)$. Now

(1)
\begin{align} \phi(u + v) = \phi(a + c, b + d) \end{align}
(2)
\begin{equation} = (a + c) - (b + d) \end{equation}
(3)
\begin{equation} = (a - b) + (c - d) \end{equation}
(4)
\begin{align} = \phi(u) + \phi(v) \end{align}

Therefore $\phi$ is a homomorphism.

The kernel of $\phi$ is found when $a = b$. That is the set $\{ (a, a) | a \in \mathbb{Z} \}$.

The set $\phi^{-1}(3)$ is found when $a - b = 3$ or when $b = a - 3$. So then $\phi^{-1}(3) = \{ (a, a - 3) | a \in \mathbb{Z} \}$.

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