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

Okay, how do I need to 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 ?

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In your question 339762, posted some 3 hours ago, you were specifically told that what you ask now is true whenever the order of the group is odd (=uneven). How come you're asking this now?! Don't you pay minimal attention to the answers you're given? –  DonAntonio Mar 24 '13 at 20:28
    
@DonAntonio I do pay very wel attention to the answers I'm given, but as I still not understand it after 3 hours of thinking, I'm now asking for a little bit more explanation. Do you have a problem with that ? –  Kasper Mar 24 '13 at 20:32
    
Of course I do: why didn't you ask for clarification in the other question? That way you could have LEARNED something and this question you could have answered it at once...! It looks like you're only posing questions, writing down answers and not thinking of them anymore.\ –  DonAntonio Mar 24 '13 at 20:34
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I think there is nothing wrong about asking this as a separate question. But it would be nice if you add a pointer where the question came up. –  azimut Mar 24 '13 at 20:34
    
@DonAntonio I would suggest analysing for 3 seconds more before you judge. If you would have done that, you could see that I have asked for clarification in that question: math.stackexchange.com/questions/339762/… But as I still didn't understand that clarification and as I thought that this as an intersting question on his own, I asked it here. –  Kasper Mar 24 '13 at 20:42

5 Answers 5

up vote 6 down vote accepted

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 $x^{-2n} = y^{-2n}$).

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Aaah, thank you very much ! But how do you know that $x$ and $y$ must have the same uneven order $2n+1$ ? –  Kasper Mar 24 '13 at 20:55
    
@Kasper This is because the order of an element divides the order of the group. $x$ and $y$ may not have the same order, but you know that $x^{|G|} = e = y^{|G|}$. –  user58514 Mar 24 '13 at 20:58
    
ooooooh of course thanks :D @rustyracketman –  Kasper Mar 24 '13 at 21:02

Another hint may be: $$x^1=x^{2m+|G|n}=y^{2m+|G|n}=y^1$$ for some integers $m$ and $n$.

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+1: Nothing wrong in applying Bezout's identity to the coprime integers $2$ and $|G|$ :-) –  Jyrki Lahtonen Mar 24 '13 at 20:28
    
+1 Nice one ... –  DonAntonio Mar 24 '13 at 20:29
    
Nice one, indeed! +1 ;-) –  amWhy Mar 25 '13 at 0:21

Hint: Since $\left| G\right|$ is odd, 2 is invertible modulo $\left| G\right|$.

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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.

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Let $a^2=b^2$ and $|G|=2k+1$ then:

$$a=a^{-|G|}a=a^{-2k}=(a^2)^{-k}=(b^2)^{-k}=b^{-2k}=b^{-|G|}b=b$$

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