Consider a group , where any element has $ord(g)=2$. Could we say that $ G \cong F_{2} $? My idea was : consider all generating elements and say $g_{1} = e_{1} \dots$ am I right?
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$F_2$? ${}{}{}$ – Andrew Tawfeek Sep 16 '17 at 08:18
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Well, $\mathbb{F}_2$ has exactly two elements. – Giovanni De Gaetano Sep 16 '17 at 08:18
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Consider the direct sum of copies of $F_2$. – Chris Gerig Sep 16 '17 at 08:18
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3There is no such group. – José Carlos Santos Sep 16 '17 at 08:21
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Any element except the identity (which has order $1$). – drhab Sep 16 '17 at 08:21
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The free group on two generators contains no element of order $2.$ – bof Sep 16 '17 at 10:01
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The groups $G$ such that $g^2=e$ for all $g\in G$ are exactly the $\Bbb F_2$-vector spaces (therefore much more stuff than your guess). Exercise for you is showing the crucial part of this: any such group $G$ must be abelian.
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No, let $(\Bbb Z_2,+)$, and consider for instance $\Bbb Z_2\times \Bbb Z_2$. Every element has order two, but $G$ has four elements. More generally the product of $n$ copies of $\Bbb Z_2$ has $2^n$ elements all of them of order $2$.
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