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what is the meaning of this $ A=\bigcup_{x\in A}\{x,x^{-1}\}. $ ?

I found the this $A$ from here https://math.stackexchange.com/a/268877/557708

jasmine
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    It states that "$A$ equals the set given by taking the union of every set of the form ${x,x^{-1}}$ where $x\in A$" – Peter Foreman Nov 03 '19 at 13:44
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    "The elements of $A$ pair up: if $x$ is in $A$ so is $x^{-1}$, and vice versa." – kimchi lover Nov 03 '19 at 13:44
  • thanks @PeterForeman now my doubt is that why does that imply that there are an odd number of elements of order $2$? – jasmine Nov 03 '19 at 13:48
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    @jasmine Clearly since any x in A is distinct from its inverse there must be an odd number of elements in A. Then in the link you gave the group G where all these elements reside has even order. Naturally the identity is the only element of order not equal to 2 which isn't in A, so for the group G to have even order you would require an odd number of order 2 elements. – Soapy Loaf Nov 03 '19 at 13:52
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    $|G|$ is even and $|A|$ is even so $|G\setminus A|$ is even but $G\setminus A$ contains exactly the identity element of order $1$ and every element of order $2$. Thus there must be an odd number of elements of order $2$. – Peter Foreman Nov 03 '19 at 13:55
  • thanks u so much @PeterForeman – jasmine Nov 03 '19 at 13:57

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