Can you give me an example of an infinite group in which every element has order $3$ (except identity) ?
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2$\Bbb Z_3 \times \Bbb Z_3\times \Bbb Z_3 \times \cdots$ – Arthur Jan 30 '15 at 12:29
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1Every finitely generated group where every element has order $3$ is finite. Which is why most of the answers involve dots...... See here for a proof. – user1729 Jan 30 '15 at 16:20
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$(\Bbb F_3[X],+){}{}{}{}{}{}$.
Marc van Leeuwen
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@DerekHolt: No, this is not isomorphic to an infinite direct product of factors $\Bbb Z/3\Bbb Z$ (Arthur's comment to the question); that would be isomorphic to $(\Bbb F_3[[X]],+)$, which is considerably larger (uncountable). My answer is indeed isomorphic to the infinite direct sum (answer posted by Dietrich Burde not long after this one). The Abelian group in the answer by P Vanchinathan is also isomorphic to this one, with a lot of choice for the isomporphism. – Marc van Leeuwen Jan 30 '15 at 13:56
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Let $A$ denote the set of positive rational numbers, an abelian group with respect to multiplication. Define $f\colon A\to A$ by $f(x)=x^3$. This is a homomorphism, is injective, but not an automorphism. For example $4$ is not in the image. SO the image subgroup $f(A)$ is a proper subgroup. The quotient group $A/f(A)$ meets your requirement. All the cosets $pf(A)$ for $p$ varying over primes are distinct, so this is an infinite group.
P Vanchinathan
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It is quotient of the group of positive rationals, hence countable. – P Vanchinathan Jan 30 '15 at 15:32
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@Marc: I read your comment ignoring the parenthesis, and interpreted it as saying "No, it is not countable", and rushed with the above comment! – P Vanchinathan Jan 30 '15 at 15:34