# element of odd finite group has unique square root

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)"

• 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? Apr 10, 2016 at 13:45
• ahow -》show i edited answer
– 김일희
Apr 10, 2016 at 14:18

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.

• 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?
– 김일희
Apr 10, 2016 at 14:38
• Correct. No other conditions are needed.
– lulu
Apr 10, 2016 at 14:40
• Ohhhh thank you thankyou very much... i have considered this problem for two days... im so stupid... haha thank you!!!!
– 김일희
Apr 10, 2016 at 14:43
• Not at all! I've been doing this for a very long time.
– lulu
Apr 10, 2016 at 14:47