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.

Prove if $g$ is an element of order $d$ and $d$ divides $n$, then $gn = 1$.

I need help on how to prove the converse which comes from this theorem that I solved.

Suppose $gn = 1$ in a group $G$ and let $d$ be the order of $g$. Then $d$ divides $n$.

Proof: By the division algorithm, we may write $n = dq + r$ with $0 < r < d$. Then

$$g^r = g^{n-dq}=g^n=(g^d)^{-q}=1 \times 1=1.$$

If r > 0, this contradicts minimality of the order $d$. Hence $r = 0$ and so $d$ divides $n$.

share|improve this question
    
Why did you suppress the things you tried to solve this? This makes for an extremely bad question. –  Did Apr 3 '13 at 17:36

2 Answers 2

Hint:

Divide $\,n\,$ by $\,d\,$ with residue:

$$n=xd+r\;,\;\;r=0\,\,\vee |r|<d\implies 1=g^n=g^{dx+r}=(g^d)^xg^r=1\cdot g^r=g^r\implies\ldots$$

share|improve this answer
    
He already proved that. He's asking about the easier converse, see his comment to my answer. –  Math Gems Apr 3 '13 at 16:56
1  
The last part of his post says "Hence r= 0 and so d divides n"...I think the OP confused the question as he just posted a few minutes ago a question asking about the direction you show here (and I also answered that other question)... –  DonAntonio Apr 3 '13 at 16:59
    
+1 for one of my teacher here. –  B. S. Apr 3 '13 at 17:14

Hint $\rm\ \color{#C00}{g^d = 1},\,\ d\mid n\:\Rightarrow\: n = dk\:\Rightarrow\: g^n = (\color{#C00}{g^d})^k = \color{#C00}1^k = 1$

Your proof of the other direction is correct. It's not clear which one you are asking about. But now you have both.

share|improve this answer
    
@9599 As I thought, see the above hint. –  Math Gems Apr 3 '13 at 16:54
    
@MathGems, the question the OP posted is a sloppy one: he did not actually ask what you're answering but what I did, as can be seen from the effort he shows in his post. Pretty confusing, indeed. –  DonAntonio Apr 3 '13 at 17:01
    
@DonAntonio No, he did. The titled question, is about "the converse of of theorem that I solved", not about the theorem "already solved". The OP confirms that in a comment above. But I can see how it might be parsed differently. –  Math Gems Apr 3 '13 at 17:05
    
The "converse part" is the most confusing, I agree...and it's fact he didn't prove anything as he asked about both directions! Anyway, read his efforts in his first post and I think it'll be obvious he meant the other direction –  DonAntonio Apr 3 '13 at 17:12
2  
The mistery's solved! Then you want Mathgems answer, which is practically the same I wrote in your other question which, even more inexplicably now after your last comment, you asked after this one...!! –  DonAntonio Apr 3 '13 at 17:17

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.