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True or False? Every infinite group has an element of infinite order.

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data

    True or False? Every infinite group has an element of infinite order.

    2. Relevant equations

    A group is a set G along with an operation * such that
    if a,b,c [itex]\in[/itex] G then
    there exists an e in G such that a*e=a
    for every a in G there exists an a' such that a*a'=e

    The order of an element is the smallest number of times it needs to be operated with itself to become equal to the identity.

    3. The attempt at a solution

    The back of the book says this is false. But I am having a hard time thinking of an infinite group where every element has finite order. Perhaps maybe the group of integers under subtraction? This is indeed a group because for a,b in Z, a-b is in Z. And a-0=a. and a-a=0 so a is its own inverse.

    So every element in Z has order 2 but the group is infinite because there are infinite integers ... Is this right?

  2. jcsd
  3. Mar 8, 2013 #2
    Consider the unit circle group of complex numbers, can you think of finite subgroups to this?
  4. Mar 8, 2013 #3


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    Z under subtraction most certainly is not a group. The operation is not necessarily associative unless you only take the positive integers.

    Speaking of the positive integers lets consider the following.

    Take the group of quotients under addition, lets call it Q+. Then the group of positive integers call it Z+ is a subgroup of the quotients under addition ( You can verify this yourself ).

    Now consider the factor group Q+/Z+. What does the most general element of this group look like? What is it's order? What is the order of Q+/Z+?
  5. Mar 8, 2013 #4
    The most general element looks like Z+q={qz: q is in Q+}? The identity of this group is just Z+, right? So the order of Z+q is the order of q? But what if q has infinite order?

    I know what I wrote above is probably wrong in many ways but I'm confused. :(
  6. Mar 8, 2013 #5


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    If you want a simpler example you could think of an infinite product of cyclic groups. I think what Zondrina is describing is basically the same thing fortissimo was hinting at. A circle group of rational points like Q/Z.
  7. Mar 8, 2013 #6
    but would they still all have the same operation? What would that operation be?

    Isn't Q/Z={qZ:q is in Q}? How are those elements of finite order? For q' being a specific element in Q isn't qZ={q+z: z is in Z} (because the operation is addition)? So how is that set of finite order? How do you operate a set with itself anyway?
  8. Mar 8, 2013 #7


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    If you don't know quotient groups then try the group consisting of the numbers ##e^{2 \pi i q}## under multiplication where q is a rational.
    Last edited: Mar 8, 2013
  9. Mar 8, 2013 #8

    So ##e^{2 \pi i qd}##=##e^{2 \pi qd}## by euler's theorem, but I don't see how that can be the identity for any natural number d? Isn't the identity of this group e^0=1?

    I was also wondering if you (or someone) could help answer my questions in the previous posts of this thread about where I am going wrong in trying to work with quotient groups?
  10. Mar 8, 2013 #9


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    That's not what Euler's theorem says. I was trying to look for an easy example you knew but I'm failing. Ok, go back to the quotient group. Sure, an element q is given by the equivalence class {q+z: z an element of Z}. That's an infinite set but it represents a single element in the quotient Q/Z. If p is another rational then it's equivalence class is {p+z: z an element of Z}. You add them by adding together the elements in the equivalence classes. So p+q={p+q+z: z an element of Z}. So notice things like 1=0 in the sense that the have the same equivalence class. Making any sense?
  11. Mar 8, 2013 #10
    Oh whoops! I was doing [itex]e^{2\pi i+pd}[/itex] rather than [itex]e^{2\pi i(pd)}[/itex].
    So what euler's theorem actually says is:

    [itex]e^{2\pi i(pd)}=cos(2(pd)\pi)+isin(2(pd)\pi)[/itex] and if we let d be the denominator of p, pd will be an integer so [itex]cos(2(pd)\pi)+sin(2(pd)\pi)=1+0=1[/itex]. so letting d be the denominator of p shows that elements in the group consisting of [itex]e^{2\pi i(p)},\hspace{5pt}p\in{\mathbb{Q}}[/itex] all have finite order (because 1 is the identity).

    Furthermore since [itex]cos((.0001)2\pi)+isin((.0001)2\pi)\neq cos((.000001)2\pi)+isin((.000001)2\pi)\neq\hspace{7pt}\text{etc. etc.}[/itex] the group consisting of [itex]e^{2\pi i(p)},\hspace{5pt}p\in{\mathbb{Q}}[/itex] is infinite. So you have indeed shown a counterexample!

    Yes that makes sense. Thanks. So all the nonnegative integers have the same equivalence class (is that the same thing as their coset being the same?), and I believe this coset is the identity of the group you have constructed, because for any rational r, we know {r+z: z is an integer}= {r+z+1: z is an integer}={r+z+2: z is an integer}={r+z+1000: z is an integer}= etc. etc. etc. (by the way, would we call the set {r+z: z is an integer} rZ or r+Z? How about the set {r+z+100: z is an integer}? Would that be rZ+100Z? Or (r+Z)+(100+Z)?)...

    The inverse for the element rZ (r+Z?) of this group (where r is an arbitrary rational) would be (-r)Z.

    Now this group has infinite order because .01Z would be different from .001Z would be different from .00000000001Z etc. etc., and if we have a coset rZ where r=a/b, then if we add rZ to itself b times, we will get the identity coset. So this means each element in our group has finite order.

    I hope this is right... If so, I have one more question... Why did we need to restrict ourselves to the positive integers? Where would negative integers have led to a problem?
    Last edited: Mar 8, 2013
  12. Mar 8, 2013 #11


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    Yes, it's right. You don't want to restrict to positive integers. I didn't. The Z in Q/Z is all integers, Zondrina was the one putting all of the pluses in. And the coset is r+Z. Not rZ. The operation here is addition. rZ isn't a coset.
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