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Let $x,y$ in a group G with odd order. Let $x^2=y^2$. Show that $x=y$.

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    $\begingroup$ Nice tagging, mate. $\endgroup$
    – Git Gud
    Nov 25, 2013 at 20:18
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    $\begingroup$ Can you show us your attempts so far? $\endgroup$
    – user93957
    Nov 25, 2013 at 20:18
  • $\begingroup$ Now that you have a good answer this comes a bit late, but if you have problems attacking a question like this my advice is to start thinking about a special case. Here you can't get smaller than $|G|=3$, which means that the group is cyclic of order three. What happens there? Can you extrapolate? $\endgroup$ Nov 25, 2013 at 20:37

1 Answer 1

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Let $k$ be such that $2k + 1$ is the order of $G$. Since $x^2 = y^2$, $(x^2)^{k+1} = (y^2)^{k+1}$.

$$ x^{2k +2} = x\cdot x^{2k+1} = y\cdot y^{2k+1} = y^{2k +2}$$

Since $2k + 1$ is the order of $G$, $x^{2k+1} = y^{2k+1} = 1_G$ and $$x = y.$$

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