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 that in an infinite cyclic group order of every element ($\ne e$) is infinite.

I have tried proving this way : If there exists an element of finite order, then it must generate a finite subgroup of $G.$ But this is not possible as by Lagrange's theorem, order of a subgroup must divide order of a group, which is not true here as order of $G$ is infinity.

Is this correct and complete ?

share|improve this question
1  
Try proving the contrapositive instead. Should be easier, now that I think about it. Then you may argue using Lagrange :) –  HowardRoark Sep 9 '13 at 17:56
add comment

4 Answers 4

up vote 2 down vote accepted

This proof is invalid, since some infinite groups have finite subgroups. E.g. $\mathbb{Z}_2 \times \mathbb{Z}$ has the subgroup $\mathbb{Z}_2 \times \{0\}$.

In an infinite cyclic group the elements are $$\{\ldots,g^{-3},g^{-2},g^{-1},1,g,g^2,g^3,\ldots,\}$$ for some generator $g$.

If $h \neq 1$ is in this group, then $h=g^k$ for some $k \in \mathbb{Z} \setminus \{0\}$. So, the subgroup generated by $h$ has the underlying set $$\{\ldots,g^{-3k},g^{-2k},g^{-k},1,g,g^{2k},g^{3k},\ldots,\}.$$

share|improve this answer
1  
ok, so as the set is infinite, order of $h$ must also be infinite. is that the logic behind ? –  Aman Mittal Sep 9 '13 at 17:51
    
That's right; the order of an element is the size of the group it generates. –  Rebecca J. Stones Sep 9 '13 at 17:53
    
ok, one side question, if a group is of infinite order, and it has a subgroup of infinite order. we call that subgroup proper or improper ? –  Aman Mittal Sep 9 '13 at 17:54
1  
A subgroup $H$ of $G$ is proper if $H \neq G$ and $H$ is not the trivial group. So, if $G$ is an infinite group, then $G$ is a subgroup of $G$ that is not proper; all other infinite subgroups would be proper. –  Rebecca J. Stones Sep 9 '13 at 17:56
    
Thanks Rebecca !! That was very helpful –  Aman Mittal Sep 9 '13 at 17:57
add comment

It doesn't quite work, as there are infinite groups with elements of (any possible) finite order.

Instead, think that every element of your infinite cyclic group $\langle g \rangle$ is of the form $x = g^k$ for some $k$. What happens if $1 = x^ n = (g^k)^n$ for some $n > 0$?

share|improve this answer
add comment

That is not correct because Lagrange's theorem presumes the group to be finite, and you can't use it that way, specifically because you're not using the fact that is cyclic.

For example $$G=\mathbb{R}-\lbrace 0 \rbrace$$

$G$ is a group under multiplication and $H=\lbrace 1,-1\rbrace$ is a finite subgroup.

You're correct trying to use negation: if a element generates a finite subgroup, the it means that $g^k=1$ for some $k$. Is that possible if the group is cyclic?

share|improve this answer
    
I am still not able to think in the right direction –  Aman Mittal Sep 9 '13 at 17:47
add comment

If you mean the infinite cyclic group $G := \{x^n | n\in \mathbb{Z}\}$. Then the proof would be by contradiction. Suppose there exists an element $y(\neq 1)\in G$ such that $y^n = 1$. But $y = x^m$ for some $m\in \mathbb{Z}$. Hence $1= y^n = (x^m)^n = x^{mn}$. But this implies G is finite, as then $G = \{x^{-mn+1},\ldots, x^{mn-1}\}$. Contradiction!

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.