# Infinite group whose every element is of order $4$?

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 $$n$$?

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