Let $G=\Bbb Z\times\Bbb Z$ be additive and $H=\{0\}\times\Bbb Z$. Show that: a) $H\le G$. b) $H\cong \Bbb Z$. I did more or less the item
a)
$H$ is a subset of $G$ so we need to show that $H$ is a group.
$H$ is associative because
$$\begin{align}
[(a, b) + (c, d)] + (e, f) &= (a + c, b + d) + (e, f) \\
&= (a + c + e, b + d + f) \\
&= (a, b) + [(c + e, d + f)] \\
&= (a, b) + [(c, d) + (e, f)].
\end{align}$$
$H$ has a neutral element because
$$(0,0) + (a, b) = (a, b)$$
Does $H$ have an inverse? (Not because $H$ is of the form $(0, x)$ where $x\in\Bbb Z$
$$(-a, -b) + (a, b) = (0.0)$$
Thank you in advance for any help.

$H $ is a subset and a group  (easy).  $\varphi:\Bbb Z\to\{0\}\times\Bbb Z $ by $x\mapsto (0,x) $ is an isomorphism.
