# Which elements generate $\mathbb Z \oplus \mathbb Z$

It's a simple question, but I have a little problem with an aspect of the proof.

I'm trying to prove that this set $\{(1,0),(0,1)\}$ generates $\mathbb Z \oplus \mathbb Z$. Following the definition 2.7 page 32 of the Hungerford's book, I have to show that every subgroup $\mathbb Z \oplus \mathbb Z$ which contains the set$\{(1,0),(0,1)\}$ must contain all of $\mathbb Z\oplus \mathbb Z$

I know also every element $(a,b)\in \mathbb Z\oplus \mathbb Z$ is written as $(a,b)=a(1,0)+b(0,1)$, I think this should take me to the conclusion, but I don't know why, I need help in this part.

EDIT

In another words, why $(a,b)\in \mathbb Z\oplus\mathbb Z\implies \{(1,0),(0,1)\}\subset \mathbb Z\oplus \mathbb Z$?

I need help here

Thanks a lot.

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Not every subgroup $H$ contains $\{(1,0),(0,1)\}$: look at $\{(0,0)\}$ or even $2\Bbb{Z}\oplus2\Bbb{Z}$. In fact, if a subgroup $H\subseteq\Bbb{Z}\oplus\Bbb{Z}$ has $\{(1,0),(0,1)\}$ as a subset, $H$ must contain $\Bbb{Z}\oplus\Bbb{Z}$, because (as you've noted) every element in $\Bbb{Z}\oplus\Bbb{Z}$ can be written as $a(1,0) + b(0,1)$ for some $a,b\in\Bbb{Z}$, and any subgroup containing some number of elements must have all multiples and combinations of multiples of those elements. – Stahl Apr 20 '13 at 3:00
@Stahl sorry, I will edit the post, you said what I meant to say – user42912 Apr 20 '13 at 3:05
@JasonDeVito sorry it's the definition 2.7 page 32 – user42912 Apr 20 '13 at 3:10
@Brad Let $G$ be a group and $X$ a subset of $G$. Let $H_i$ A family of all subgroups of $G$ which contain $X$. Then $\cap H_i$ is called the subgroup of G generated by the set X. – user42912 Apr 20 '13 at 3:13

Let $G$ be a group (written multiplicatively). If $g,h\in G$, then $$\displaystyle\prod_{i = 1}^{n} g^{k_i}h^{l_i}\in G$$ for all $n\in\Bbb{N}$, $k_i,l_i\in\Bbb{Z}$. But in your case, $\Bbb{Z}\oplus\Bbb{Z}$ is abelian, so we can do even better: let $H\subseteq\Bbb{Z}\oplus\Bbb{Z}$ be a subgroup of $\Bbb{Z}\oplus\Bbb{Z}$, and suppose $g,h\in H$. Then $ng + mh\in H$ for all $n,m\in\Bbb{Z}$. If you know this already or can show this, the solution should follow easily from your observation that any element can be written as $a(1,0) + b(0,1)$ for some $a,b\in\Bbb{Z}$.