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.

Could someone please help me answer this question? Using the subgroup test it was proved that a subgroup $H$ $\le $ $\mathbb{Z}$ is necessarily of the form ${n}$ $\bullet$ $\mathbb{Z}$ for some n $\ge$ $0$.

share|improve this question
4  
Please avoid posting the question only in the title. The body should be self-contained. –  Asaf Karagila Feb 2 '13 at 3:19
add comment

5 Answers

up vote 6 down vote accepted

HINT: Let $G$ be a subgroup of $\Bbb Z$. Let $a$ be the smallest positive element of $G$. Prove that $G=\{na:n\in\Bbb Z\}=\langle a\rangle$, the subgroup of $\Bbb Z$ generated by $a$. You may find the division algorithm useful.

share|improve this answer
    
@ Brian M.Scott Thank you –  Bulou Duikoro Feb 2 '13 at 5:13
    
@Bulou: You’re welcome. –  Brian M. Scott Feb 2 '13 at 5:40
add comment

$\Bbb Z$ is cyclic since $\Bbb Z=\langle 1\rangle$, and any subgroup of a cyclic group is cyclic.

share|improve this answer
add comment

Hint $\ $ From Euclid, a nonempty subset $\rm\,S\subseteq \Bbb Z$ closed under subtraction is closed under mod, so closed under gcd. Thus the least positive $\rm\,d\in S\,$ divides every $\rm\,n\in S\:$ (else $\rm\:gcd(d,n) < d).$

Remark $\ $ One can skip the intermediate "closed under mod" if one is familiar with the "subtractive" form of the Euclidean algorithm, which calculates the gcd by repeatedly subtracting the smaller integer from the larger. Then it is crystal clear that subgroups, being closed under subtraction, are also closed under the gcd operation.

share|improve this answer
add comment

Assume $a$ is the smallest non-zero element of the subgroup A(we can always choose the smallest one(why?)).

Then, consider any element $x\in A$,then $\exists n,r\in\Bbb Z$ such that $x=na+r$ where $0\leq r\lt a$(division algo.)

But then, $r=x-na\in A$ and $r\lt a$ which contradicts that $a$ is the smallest non-zero element unless $r=0$ itself which gives $x=na$ where $n\in\Bbb Z,x\in A$

share|improve this answer
    
@ Avatar $\mathbb{a}$ is the smallest non -zero element of the subgroup because it has the identity. –  Bulou Duikoro Feb 7 '13 at 1:12
add comment

$\quad$$\mathbb{G}$ = { $\mathbb {na}$ : $\mathbb{n}$ $\in$ $\mathbb{Z}$} = $\lt$ $\mathbb{a}$$\gt$

Let $g_1$ $\in$ $\mathbb{G}$

Now want to show that $g_1$ is in { $\mathbb {na}$ : $\mathbb{n}$ $\in$ $\mathbb{Z}$}

$\qquad$ $g_1$ = $\mathbb{k\cdot a}$ + $\mathbb{r}$ $\qquad$ $\mathbb{a}$ $\gt$ $\mathbb{r}$ $\ge$ 0

$\qquad$$\qquad$$\mathbb{r}$ = $g_1$ - $\mathbb{k\cdot a}$ $\qquad$( $\mathbb{r}$ $\in$ $\mathbb{G}$ )

therefore it was found that $\quad$$\mathbb{r}$ = 0

$\qquad$$\qquad$ $g_1$ = $\mathbb{k\cdot a}$

hence this completes the proof

share|improve this answer
add comment

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.