Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

up vote 1 down vote accepted

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 $a,b\in\Bbb{Z}$.

share|improve this answer
    
yes yes of course, so stupid question, thank you for helping me. –  user42912 Apr 20 '13 at 3:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.