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.

Let $C$ be a cyclic group. Let $A$ and $B$ be two subgroups of $C$ with $|A|=|B|$. Then $A = B$.

How to show this? Thanks.

(Btw, I already know $A$ and $B$ are cyclic)

share|improve this question
2  
This will only hold if $C$ is a finite cyclic group. E.g., $|2\mathbb Z| = |4\mathbb Z|$ but these subgroups of $\mathbb Z$ groups are not equal. –  amWhy May 1 '13 at 18:42
    
@amWhy, of course if one reformulates the question in terms of indices, then it works for the infinite cyclic group as well. –  Andreas Caranti May 1 '13 at 18:46
    
I agree, @AndreasCaranti –  amWhy May 1 '13 at 18:47
add comment

3 Answers 3

Recall that for every divisor of the order of a cyclic group $\exists$ a unique subgroup of that order. Now if $A$ and $B$ are two subgroups of a finite cyclic group $G$ then their orders must be divisors of the order of the group and due to uniqueness they must be the same.

share|improve this answer
1  
I think the question would be boringly trivial if the OP knew/could use this theorem... –  DonAntonio May 1 '13 at 19:08
    
i agree :) @donantonio –  wanderer May 1 '13 at 21:45
add comment

Hint: As amWhy noted you need $C$ to be finite, so you can assume $C = \mathbb Z/n$ for some $n$. Let $a$ be the least positive integer contained in $A$. Show that $a \mid n$ and consequently $a$ generates $A$. Then $a = n/|A|$ is uniquely identified by the size of $A$, therefore $A$ is uniquely identified by it's size.

share|improve this answer
add comment

Theorem 1: If $G=\langle a\rangle$ be a finite group of order $n$ and $$d_1,d_2,...,d_k$$ be all distinct positive divisors of $n$ so the following subgroups are all the proper distinct subgroups of $G$: $$\langle a^{d_1}\rangle,\langle a^{d_2}\rangle,...,\langle a^{d_k}\rangle $$

Theorem 2: If $G=\langle a\rangle$ be an infinite group then the following subgroups are all the proper distinct subgroups of $G$: $$\langle e_G\rangle,\langle a\rangle,\langle a^{2}\rangle,...,\langle a^{k}\rangle,... $$

share|improve this answer
    
+1 Nice to have stated explicitly the theorem we're looking at! –  amWhy May 3 '13 at 0:08
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.