I have constructed infinite group whose every element is of prime order by taking the set as set of sequences whose elements are from integers modulo $p$ and operation is integers modulo $p$.

Now how can I get an infinite group whose every element is of order $4$ (non prime) except identity?.

Is there any general way of finding an infinite group whose every element is of order $n$?

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    $\begingroup$ The square of an order $4$ element has order $2$. $\endgroup$ – Lord Shark the Unknown Sep 28 '19 at 4:22
  • $\begingroup$ In what group? You are taking? $\endgroup$ – Mr.Multitalented Sep 28 '19 at 4:24
  • $\begingroup$ Related. $\endgroup$ – Shaun Sep 28 '19 at 4:39
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    $\begingroup$ You can't do such for a non prime order. If $n = jk$ then if $|a| = jk$ then $(a^j)^k= e$ and $|a^j| \le k$. $\endgroup$ – fleablood Sep 28 '19 at 4:44

Suppose $\lvert a\rvert=4$ for some $a\in G$ for some group $G$. Then $(a^2)^2=e$ and $a^2$ is non-trivial, so $$\lvert a^2\rvert=2.$$

  • $\begingroup$ I didn't understand, do you got the question correctly,? I may not be clear in my question? We need to find a group whose every element is of order 4? $\endgroup$ – Mr.Multitalented Sep 28 '19 at 4:31
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    $\begingroup$ Yes, @Mr.Multitalented, but if an element, any element, of a group, any group, has order four, then its square has order two, as proven above; thus no group exists in which every element has order four. $\endgroup$ – Shaun Sep 28 '19 at 4:34
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    $\begingroup$ @mr.multitalented this answer is telling you that it’s imposible find the group you’re looking for. $\endgroup$ – Alonso Delfín Sep 28 '19 at 4:34
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    $\begingroup$ Okay.. Got it.. Thank you 😃 $\endgroup$ – Mr.Multitalented Sep 28 '19 at 17:35

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