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Does there exist any infinite group $G$ where each element is of order $2$?

It is clear that the group should be abelian. I tried to find out a suitable example to meet my purpose. But I have not yet found it. I think it is out of my knowledge.

So, please help.

Shaun
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    There are infinitely many $C_2\times C_2\times C_2\times\dots\times C_2$. Note that the identity must be order 1. The simplest are $C_2$ the cyclic group of order 2 and the "Klein group", for which see https://en.wikipedia.org/wiki/Klein_four-group – almagest Jun 19 '16 at 10:20
  • Do you mean that the group what you have mentioned here is of infinite order since infinitely many direct products are possible among $C_2$ to make it an infinite having the property that each non-identity element is of order 2. –  Jun 19 '16 at 10:35
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    Please use a more descriptive title. "Problem", "solving" and "question" are all words that would fit any question on the site. – joriki Jun 19 '16 at 10:38
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    Let $E$ be an infinite set. Let $\mathcal P(E)={X:X\subseteq E}.$ For $X,Y\in\mathcal P(E)$ the symmetric difference of $X$ and $Y$ is $X\triangle Y=(X\setminus Y)\cup(Y\setminus X).$ Then $(\mathcal P(E),\triangle)$ is an infinite group in which each element (except the identity) is of order $2.$ – bof Jun 19 '16 at 10:53

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The following summarises the example by @bof in the comments . . .

Let $E$ be any infinite group. Consider the power set $\mathcal{P}(E)=\{X\mid X\subseteq E\}$.

For $X, Y\in \mathcal{P}(E)$, the symmetric difference $X\vartriangle Y$ is given by $$X\vartriangle Y=(X\setminus Y)\cup(Y\setminus X).$$

Theorem: $(\mathcal{P}(E), \vartriangle)$ is an infinite group in which every non-identity element has order $2$.

Proof:

If $\mathcal{P}(E)$ were finite, so then must be $E$. Thus $\mathcal{P}(E)$ is infinite.

A proof that $(\mathcal{P}(E), \vartriangle)$ is a group with identity $\varnothing $ can be found here.

Finally, consider $\varnothing \neq A\in\mathcal{P}(E)$. Then $$\begin{align} A\vartriangle A&=(A\setminus A)\cup(A\setminus A)\\ &=\varnothing \cup \varnothing\\ & =\varnothing. \end{align}$$ Hence the order of each non-identity element of $(\mathcal{P}(E), \vartriangle)$ has order two.$\square$

Shaun
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