Let $x,y$ in a group G with uneven order. Let $x^2=y^2$. Show that $x=y$. Let $x,y$ in a group G with uneven order. Let $x^2=y^2$. Show that $x=y$.
Okay, how do I prove this? 
If $G$ is unenven order, then there are no elements of order $2$. So $x^2=y^2=e$ only if $x=e$. 
But what if the order of $x,y ≥3$ ? Someone told me that I can prove that $x\mapsto x^2$ is a bijection if $x$ has uneven order. How can I see this ?
 A: Another hint may be: $$x^1=x^{2m+|G|n}=y^{2m+|G|n}=y^1$$ for some integers $m$ and $n$.
A: The order of the group $G$ is odd, so it can be written as $2n + 1$ for some $n$. 
Since the order of any element divides the order of the group, you have $1 = x^{2n + 1} = y^{2n + 1}$. You also have $x^{2n} = y^{2n}$ since $x^2 = y^2$. Since $x^{2n}x = y^{2n}y$, this means $x$ and $y$ have the same inverse, which in turn means $x = y$. (Or just multiply through by $x^{-2n} = y^{-2n}$).
A: Hint: Since $\left| G\right|$ is odd, 2 is invertible modulo $\left| G\right|$.
A: Hint $\rm\ \ \ \color{#C00}{X^J\!=Y^J},\, \ \color{#0A0}{X^K\! = Y^K}\Rightarrow\ X^{(J,K)}\! = Y^{(J,K)},\ $ where $\rm\:(J,K)\:$ denotes $\rm\:gcd(J,K).$
Proof $\ $ By Bezout's gcd identity $\rm\ (J,K)\, =\, mJ\!-\!nK\ $ for some $\rm\ m,n\in\Bbb Z.\:$ Therefore
$$\rm X^{(J,K)} = X^{mJ-nK}\! = (\color{#C00}{X^J})^m (\color{#0A0}{X^K})^{-n} = (\color{#C00}{Y^J})^m (\color{#0A0}{Y^K})^{-n} = Y^{mJ-nK}\! = Y^{(J,K)}\quad {\bf QED}$$
Remark $\ $ Yours is the special case $\rm\:J = 2^n,\:$ $\rm\,K\,$ odd (= group order), $ $ so the gcd $\rm\:(J,K) = 1.$
Or, more conceptually, the set of exponents $\rm\:n\:$ such that $\rm\:X^n\! = Y^n$ are closed under subtraction so closed under gcd, since they form a subgroup/ideal of $\rm\,\Bbb Z,\,$ which is necessarily cyclic/principal, generated by the gcd of all its elements.
A: Let $a^2=b^2$ and $|G|=2k+1$ then:
$$a=a^{-|G|}a=a^{-2k}=(a^2)^{-k}=(b^2)^{-k}=b^{-2k}=b^{-|G|}b=b$$
