Let $G=\Bbb Z\times\Bbb Z$ be additive and $H=\{0\}\times\Bbb Z$. Show that: a) $H\le G$. b) $H\cong \Bbb Z$.

Question

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.

Answer 1

For a)

I will use the one-step subgroup test.

Since $(0,0)\in H$, we have $H\neq \varnothing$.

By definition of $H$, we have $H\subseteq G$.

Let $x,y\in H$. Then $x=(0,a), y=(0,b)$ for some $a,b\in\Bbb Z$. Now

$$\begin{align} x-y&=(0,a)-(0,b)\\ &=(0,a)+(0,-b)\\ &=(0,a-b). \end{align}$$

But $a-b\in\Bbb Z$, so $x-y\in H$.

Hence $H\le G$.

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For b), consider

$$\begin{align} \varphi: H&\to \Bbb Z\\ (0,h)&\mapsto h. \end{align}$$

Answer 2

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