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let G be an odd finite group show that each element of G has a unique square root (if $x \in G$ then there exists $y \in G$ such that $x=y^2$)

Is this problem right? If there is another condition such as G is commutative, I can solve this problem But with only this, My logic fails. My question is that "do this problem without condition: abelian, make sense?? Is there any counterexamples? (odd finite group but there exists an element which doesn't have square root)"

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    $\begingroup$ What is the problem? Are you asking if there is a group with an odd number of elements such that each element has a unique square root? $\endgroup$
    – almagest
    Commented Apr 10, 2016 at 13:45
  • $\begingroup$ ahow -》show i edited answer $\endgroup$
    – 김일희
    Commented Apr 10, 2016 at 14:18

1 Answer 1

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First we prove that the square root (if it exists) is unique: If $\#G=2k-1$ and $g_1^2=g_2^2$ then we write $$g_1=g_1^{2k}=(g_1^2)^{k}=(g_2^2)^k=g_2^{2k}=g_2$$

Now we remark that this suffices. Indeed, if $f:S\to S$ is a function from a finite set to itself, then injectivity is equivalent to surjectivity.

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  • $\begingroup$ Thanks for your answer!! (: you say that your function f is bijective and therefore all elements of G have the unique square root. And so the problem don't need another condition. Right??? Did i understand you correctly? $\endgroup$
    – 김일희
    Commented Apr 10, 2016 at 14:38
  • $\begingroup$ Correct. No other conditions are needed. $\endgroup$
    – lulu
    Commented Apr 10, 2016 at 14:40
  • $\begingroup$ Ohhhh thank you thankyou very much... i have considered this problem for two days... im so stupid... haha thank you!!!! $\endgroup$
    – 김일희
    Commented Apr 10, 2016 at 14:43
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    $\begingroup$ Not at all! I've been doing this for a very long time. $\endgroup$
    – lulu
    Commented Apr 10, 2016 at 14:47

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