0
$\begingroup$

Essentially, i need this for a third year mathematics project and originally i thought i just needed to have something like this:

The group with this presentation is explicitly realized by the set of integers , under the operation of addition, where $a = 1$. Hence there exists an isomorphism from the infinite cyclic group of order $n$ to the integers mod $n$.

But i was told that i actually needed to show that there exists a free group that is generated by $a$, where $a$ is an element of the infinite cyclic group. Thank you.

$\endgroup$
  • 1
    $\begingroup$ Could you please state your question precisely? From the title it looks like you are trying to prove that every infinite cyclic group can be presented with one generator and no relations. Is this correct? If so, then the reason your proof was rejected is because you are proving the converse statement, i.e. that a group with presentation $\langle a \mid \rangle$ is cyclic of infinite order. $\endgroup$ – A.P. Jan 7 '16 at 22:50
  • $\begingroup$ Hi. yes, i'm trying to prove that every infinite cyclic group can be presented with one generator and no relations. And you are right, that's what my supervisor suggested when i asked why the proof was rejected. Do you know how to prove the original statement? Thank you. $\endgroup$ – Abisola Hammed Jan 8 '16 at 23:22
1
$\begingroup$

A cyclic group by definition is one generated by one of its elements. This means that the set of elements $a^1, a^2, a^3, ...$ gives all elements of the group. For a finite group this implies that $a^n = 1$ (the identity element) for some n, n being the number of elements in the group (the group order).

For infinite cyclic groups the situation is similar, except that $a^0 = 1$ is added to the list, and $a^1, a^2, a^3, ...$ and$a^{-1}, a^{-2}, a^{-3}, ...$have to form the set of elements unequal to 1. So yes you need to find an element a that generates all group members. Note that $a^n$ may be read as $a+a+a+...+a$ (n times) for groups with additive + operation.

$\endgroup$
  • $\begingroup$ You forgot the inverses $a^{-1}, a^{-2}, ...$. $\endgroup$ – Paul K Jan 7 '16 at 22:55
  • $\begingroup$ you're absolutely right - I will add these - thx $\endgroup$ – Maestro13 Jan 7 '16 at 23:03
  • $\begingroup$ Thank you for your help. But whilst i understand your definition of a cyclic group of infinite order, i don't really get how i can use that to prove that there exists a free group that is generated by this generator (element). $\endgroup$ – Abisola Hammed Jan 8 '16 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.