Let G be a group generated out of a set of $n$ distinct elements such that each of them has order $2$ and furthermore each element of the group has order $2$.

The group acts as follows. $a*b=a*b\neq b*a$ however $a*b*a*b$ is the identity since all elements have order 2.

When there are only $2$ elements $a,b$ this set has size $7: e,a,b,ab,ba,aba,bab$

How can I find the size for a given $n$?

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    $\begingroup$ If every element in a group has order two, the group is necessarily abelian. $1=aabb=abab\implies ab=ba$. $\endgroup$ – Pedro Tamaroff Apr 2 '14 at 3:36
  • $\begingroup$ Oh, so then this group is simply the product $(Z/2)^2$? $\endgroup$ – Jorge Fernández Hidalgo Apr 2 '14 at 4:29
  • $\begingroup$ What happens if we let them have order 3? $\endgroup$ – Jorge Fernández Hidalgo Apr 2 '14 at 4:32
  • $\begingroup$ if each element has order $2$, such a group has form $Z_2\times Z_2\times ..$ so order of group is $2^n$ if $n$ is the minimal number of generators $\endgroup$ – mesel Apr 2 '14 at 6:24
  • $\begingroup$ @mesel thanks, I got that $\endgroup$ – Jorge Fernández Hidalgo Apr 2 '14 at 20:05

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