Group Homomorphisms and Kernels Consider the group $\mathbb{Z}^+$. Define $f : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ by $f(x,y) = x+y$. Show that $f$ is a homomorphism and find the kernel.
Detirmine whether the kernel is a cyclic subgroup.
 A: To be a homomorphism the function $f:\mathbb Z\times \mathbb Z\to \mathbb Z$ must hold $ f(u+v) = f(u) + f(v) $ Let's say that $u_1, u_2, v_1, v_2\in \mathbb Z$. Then $(u_1,v_1), (u_2, v_2)\in \mathbb Z \times \mathbb Z$ and $(u_1,v_1)+(u_2, v_2)=(u_1+u_2,v_1+v_2)\in \mathbb Z\times \mathbb Z$. Since $f[(u_1,v_1)+(u_2, v_2)]= f(u_1+u_2,v_1+v_2)=u_1+u_2+v_1+v_2=f(u_1,v_1)+f(u_2, v_2)$. So, it is trivial that $f$ is a homomorphism.
Kernel is defined as $\mathrm{ker}(f) := \{u \in \mathbb Z \times \mathbb Z : f(u) = e_{\mathbb Z}\}$. Since $e_{\mathbb Z}$ is $0$, $\mathrm{ker}(f)$ is $\{(u,-u)|u\in \mathbb Z\}$. Since identity of $\mathrm{ker}(f)$ is $(0,0)$ and $\mathrm{ker}(f)$ is generated from $(1,-1)$ and $(-1,1)$ is its inverse element it is a infinite cyclic subgroup.
A: You know that $\Bbb{Z}^2$ forms a group.  So you only need to check whether $f: \Bbb{Z}^2 \to \Bbb{Z}$ is a homomorphism.  It's a homomorphism $\iff$ it preserves structure:
$f(a + b) = f(a) + f(b),  \forall a,b \in \Bbb{Z}^2$
Prove that. Then the kernel of $f$ is the part of the domain on which it vanishes.  So setting the formula $f(x, y) = x+y$ equal to zero and solving for $(x,y)$ will give you an identification of the kernel.
We have: $x + y = 0 \iff x = -y \iff (x,y) \in \{(r,-r) : r \in \Bbb{Z}\}$.  Thus $\ker f = $ the line through $0$ and $(1, -1)$ in $\Bbb{Z}^2$.  It's obviously cyclic as it equals $\Bbb{Z}\cdot(1,-1)$.
For example $\Bbb{Z}/n\Bbb{Z} = \Bbb{Z}_n$ is cyclic as it equals $\Bbb{Z}\cdot \{1\} \pmod n = \{\dots, (-1 + -1), -1, 0, 1, (1+1), \dots \}$, or said yet another way, there exist an element $e\in G$, the group, such that $G = \{\dots, 1(-e), 0, 1e, \dots \}$.  That's three ways of saying that it's cyclic, it basically means it's a $\Bbb{Z}$-module of dimension $1$.  There's another way.
