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Let there be an Abelian group with a binary operation $\ast$ on a set $S$. Let such a group respect the following propriety:

$$ (X\ast Y)\ast Y = X$$

For any $X$ and $Y$ in $S$. I realize that by using the associative propriety we can say:

$$ (X\ast Y)\ast Y = X \;\therefore\; X\ast (Y\ast Y) = X \;\therefore\; Y\ast Y = e$$

Where $e$ is the identity element of $\ast$, meaning that under $\ast$, any element of $S$ is it's own inverse. Could such a group exist? If so, what kind of proprieties would it have? Are there any examples?

Much appreciated.

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2  
Addition modulo $2$. Many others. –  André Nicolas Apr 14 at 17:05

2 Answers 2

up vote 2 down vote accepted

Yes, there are many groups where every element satisfies $Y * Y = e$. The simplest one is $\mathbb{Z}/2\mathbb{Z}$, the group of integers modulo 2 (under addition). These groups have interesting properties. For example you don't have to assume that the group is abelian; if a group is such that $Y * Y = e$ for all $Y$, it's automatically abelian.

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And is there a name for such a group? –  Disousa Apr 14 at 19:04
1  
They're called "groups with exponent two". –  Najib Idrissi Apr 14 at 20:57

Since it must hold for every element, your condition is equivalent to the following statement: any non-identity element has order two.

Andre's example is such an example, as is the group $\{-1,1\}$ with multiplication.

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What do you mean by "any non-identity element has order two"? –  Disousa Apr 14 at 19:21
    
If $x\neq e$ (where $e$ is the identity), then $x^2=e$. –  Hayden Apr 14 at 22:38

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